Rd Sharma Xi 2020 2021 _volume 1 Solutions for Class 11 Science Maths Chapter 18 Binomial Theorem are provided here with simple step-by-step explanations. These solutions for Binomial Theorem are extremely popular among Class 11 Science students for Maths Binomial Theorem Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 2021 _volume 1 Book of Class 11 Science Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 2021 _volume 1 Solutions. All Rd Sharma Xi 2020 2021 _volume 1 Solutions for class Class 11 Science Maths are prepared by experts and are 100% accurate.

Question 1:

Using binomial theorem, write down the expansions of the following:
(i) ${\left(2x+3y\right)}^{5}$

(ii) ${\left(2x-3y\right)}^{4}$

(iii) ${\left(x-\frac{1}{x}\right)}^{6}$

(iv) ${\left(1-3x\right)}^{7}$

(v) ${\left(ax-\frac{b}{x}\right)}^{6}$

(vi) ${\left(\frac{\sqrt{x}}{a}-\sqrt{\frac{a}{x}}\right)}^{6}$

(vii) ${\left(\sqrt[3]{x}-\sqrt[3]{a}\right)}^{6}$

(viii) ${\left(1+2x-3{x}^{2}\right)}^{5}$

(ix) $\left(x+1-\frac{1}{x}\right)$

(x) ${\left(1-2x+3{x}^{2}\right)}^{3}$

(i) (2x + 3y)5

$={}^{5}C_{0}\left(2x{\right)}^{5}\left(3y{\right)}^{0}+{}^{5}C_{1}\left(2x{\right)}^{4}\left(3y{\right)}^{1}+{}^{5}C_{2}\left(2x{\right)}^{3}\left(3y{\right)}^{2}+{}^{5}C_{3}\left(2x{\right)}^{2}\left(3y{\right)}^{3}+{}^{5}C_{4}\left(2x{\right)}^{1}\left(3y{\right)}^{4}+{}^{5}C_{5}\left(2x{\right)}^{0}\left(3y{\right)}^{5}$
$=32{x}^{5}+5×16{x}^{4}×3y+10×8{x}^{3}×9{y}^{2}+10×4{x}^{2}×27{y}^{3}+5×2x×81{y}^{4}+243{y}^{5}\phantom{\rule{0ex}{0ex}}=32{x}^{5}+240{x}^{4}y+720{x}^{3}{y}^{2}+1080{x}^{2}{y}^{3}+810x{y}^{4}+243{y}^{5}\phantom{\rule{0ex}{0ex}}$

(ii) (2x − 3y)4

$={}^{4}C_{0}\left(2x{\right)}^{4}\left(3y{\right)}^{0}-{}^{4}C_{1}\left(2x{\right)}^{3}\left(3y{\right)}^{1}+{}^{4}C_{2}\left(2x{\right)}^{2}\left(3y{\right)}^{2}-{}^{4}C_{3}\left(2x{\right)}^{1}\left(3y{\right)}^{3}+{}^{4}C_{4}\left(2x{\right)}^{0}\left(3y{\right)}^{4}\phantom{\rule{0ex}{0ex}}=16{x}^{4}-4×8{x}^{3}×3y+6×4{x}^{2}×9{y}^{2}-4×2x×27{y}^{3}+81{y}^{4}\phantom{\rule{0ex}{0ex}}=16{x}^{4}-96{x}^{3}y+216{x}^{2}{y}^{2}-216x{y}^{3}+81{y}^{4}$

(iii)

(iv) (1 − 3x)7
$={}^{7}C_{0}\left(3x{\right)}^{0}-{}^{7}C_{1}\left(3x{\right)}^{1}+{}^{7}C_{2}\left(3x{\right)}^{2}-{}^{7}C_{3}\left(3x{\right)}^{3}+{}^{7}C_{4}\left(3x{\right)}^{4}-{}^{7}C_{5}\left(3x{\right)}^{5}+{}^{7}C_{6}\left(3x{\right)}^{6}-{}^{7}C_{7}\left(3x{\right)}^{7}\phantom{\rule{0ex}{0ex}}=1-7×3x+21×9{x}^{2}-35×27{x}^{3}+35×81{x}^{4}-21×243{x}^{5}+7×729{x}^{6}-2187{x}^{7}\phantom{\rule{0ex}{0ex}}=1-21x+189{x}^{2}-945{x}^{3}+2835{x}^{4}-5103{x}^{5}+5103{x}^{6}-2187{x}^{7}$

(v)

$\left(ax-\frac{b}{x}{\right)}^{6}\phantom{\rule{0ex}{0ex}}={}^{6}C_{0}\left(ax{\right)}^{6}\left(\frac{b}{x}{\right)}^{0}-{}^{6}C_{1}\left(ax{\right)}^{5}\left(\frac{b}{x}{\right)}^{1}+{}^{6}C_{2}\left(ax{\right)}^{4}\left(\frac{b}{x}{\right)}^{2}-{}^{6}C_{3}\left(ax{\right)}^{3}\left(\frac{b}{x}{\right)}^{3}+{}^{6}C_{4}\left(ax{\right)}^{2}\left(\frac{b}{x}{\right)}^{4}-{}^{6}C_{5}\left(ax{\right)}^{1}\left(\frac{b}{x}{\right)}^{5}+{}^{6}C_{6}\left(ax{\right)}^{0}\left(\frac{b}{x}{\right)}^{6}$
$={a}^{6}{x}^{6}-6{a}^{5}{x}^{5}×\frac{b}{x}+15{a}^{4}{x}^{4}×\frac{{b}^{2}}{{x}^{2}}-20{a}^{3}{b}^{3}×\frac{{b}^{3}}{{x}^{3}}+15{a}^{2}{x}^{2}×\frac{{b}^{4}}{{x}^{4}}-6ax×\frac{{b}^{5}}{{x}^{5}}+\frac{{b}^{6}}{{x}^{6}}\phantom{\rule{0ex}{0ex}}={a}^{6}{x}^{6}-6{a}^{5}{x}^{4}b+15{a}^{4}{x}^{2}{b}^{2}-20{a}^{3}{b}^{3}+15\frac{{a}^{2}{b}^{4}}{{x}^{2}}-6\frac{a{b}^{5}}{{x}^{4}}+\frac{{b}^{6}}{{x}^{6}}$

(vi)
${\left(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}\right)}^{6}\phantom{\rule{0ex}{0ex}}={}^{6}C_{0}{\left(\sqrt{\frac{x}{a}}\right)}^{6}{\left(\sqrt{\frac{a}{x}}\right)}^{0}-{}^{6}C_{1}{\left(\sqrt{\frac{x}{a}}\right)}^{5}{\left(\sqrt{\frac{a}{x}}\right)}^{1}+{}^{6}C_{2}{\left(\sqrt{\frac{x}{a}}\right)}^{4}{\left(\sqrt{\frac{a}{x}}\right)}^{2}-{}^{6}C_{3}{\left(\sqrt{\frac{x}{a}}\right)}^{3}{\left(\sqrt{\frac{a}{x}}\right)}^{3}+{}^{6}C_{4}{\left(\sqrt{\frac{x}{a}}\right)}^{2}{\left(\sqrt{\frac{a}{x}}\right)}^{4}-{}^{6}C_{5}{\left(\sqrt{\frac{x}{a}}\right)}^{1}{\left(\sqrt{\frac{a}{x}}\right)}^{5}+{}^{6}C_{6}{\left(\sqrt{\frac{x}{a}}\right)}^{0}{\left(\sqrt{\frac{a}{x}}\right)}^{6}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{3}}{{a}^{3}}-6\frac{{x}^{2}}{{a}^{2}}+15\frac{x}{a}-20+15\frac{a}{x}-6\frac{{a}^{2}}{{x}^{2}}+\frac{{a}^{3}}{{x}^{3}}\phantom{\rule{0ex}{0ex}}$

(vii)
${\left(\sqrt[3]{x}-\sqrt[3]{a}\right)}^{6}\phantom{\rule{0ex}{0ex}}={}^{6}C_{0}\left(\sqrt[3]{x}{\right)}^{6}\left(\sqrt[3]{a}{\right)}^{0}-{}^{6}C_{1}\left(\sqrt[3]{x}{\right)}^{5}\left(\sqrt[3]{a}{\right)}^{1}+{}^{6}C_{2}\left(\sqrt[3]{x}{\right)}^{4}\left(\sqrt[3]{a}{\right)}^{2}-{}^{6}C_{3}\left(\sqrt[3]{x}{\right)}^{3}\left(\sqrt[3]{a}{\right)}^{3}+{}^{6}C_{4}\left(\sqrt[3]{x}{\right)}^{2}\left(\sqrt[3]{a}{\right)}^{4}-{}^{6}C_{5}\left(\sqrt[3]{x}{\right)}^{1}\left(\sqrt[3]{a}{\right)}^{5}+{}^{6}C_{6}\left(\sqrt[3]{x}{\right)}^{0}\left(\sqrt[3]{a}{\right)}^{6}\phantom{\rule{0ex}{0ex}}={x}^{2}-6{x}^{5/3}{a}^{1/3}+15{x}^{4/3}{a}^{2/3}-20xa+15{x}^{2/3}{a}^{4/3}-6{x}^{1/3}{a}^{5/3}+{a}^{2}$

(viii)

(ix)
$\left(x+1-\frac{1}{x}{\right)}^{3}\phantom{\rule{0ex}{0ex}}={}^{3}C_{0}\left(x+1{\right)}^{3}\left(\frac{1}{x}{\right)}^{0}-{}^{3}C_{1}\left(x+1{\right)}^{2}\left(\frac{1}{x}{\right)}^{1}+{}^{3}C_{2}\left(x+1{\right)}^{1}\left(\frac{1}{x}{\right)}^{2}-{}^{3}C_{3}\left(x+1{\right)}^{0}\left(\frac{1}{x}{\right)}^{3}$
$=\left(x+1{\right)}^{3}-3\left(x+1{\right)}^{2}×\frac{1}{x}+3\frac{x+1}{{x}^{2}}-\frac{1}{{x}^{3}}\phantom{\rule{0ex}{0ex}}={x}^{3}+1+3x+3{x}^{2}-\frac{3{x}^{2}+3+6x}{x}+3\frac{x+1}{{x}^{2}}-\frac{1}{{x}^{3}}\phantom{\rule{0ex}{0ex}}={x}^{3}+1+3x+3{x}^{2}-3x-\frac{3}{x}-6+\frac{3}{x}+\frac{3}{{x}^{2}}-\frac{1}{{x}^{3}}\phantom{\rule{0ex}{0ex}}={x}^{3}+3{x}^{2}-5+\frac{3}{{x}^{2}}-\frac{1}{{x}^{3}}$

(x)
$\left(1-2x+3{x}^{2}{\right)}^{3}\phantom{\rule{0ex}{0ex}}={}^{3}C_{0}\left(1-2x{\right)}^{3}+{}^{3}C_{1}\left(1-2x{\right)}^{2}\left(3{x}^{2}\right)+{}^{3}C_{2}\left(1-2x\right)\left(3{x}^{2}{\right)}^{2}+{}^{3}C_{3}\left(3{x}^{2}{\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(1-2x{\right)}^{3}+9{x}^{2}\left(1-2x{\right)}^{2}+27{x}^{4}\left(1-2x\right)+27{x}^{6}\phantom{\rule{0ex}{0ex}}=1-8{x}^{3}+12{x}^{2}-6x+9{x}^{2}\left(1+4{x}^{2}-4x\right)+27{x}^{4}-54{x}^{5}+27{x}^{6}\phantom{\rule{0ex}{0ex}}=1-8{x}^{3}+12{x}^{2}-6x+9{x}^{2}+36{x}^{4}-36{x}^{3}+27{x}^{4}-54{x}^{5}+27{x}^{6}\phantom{\rule{0ex}{0ex}}=1-6x+21{x}^{2}-44{x}^{3}+63{x}^{4}-54{x}^{5}+27{x}^{6}\phantom{\rule{0ex}{0ex}}$

Question 2:

Evaluate the following:
(i) ${\left(\sqrt{x+1}+\sqrt{x-1}\right)}^{6}+{\left(\sqrt{x+1}-\sqrt{x-1}\right)}^{6}$

(ii) ${\left(x+\sqrt{{x}^{2}-1}\right)}^{6}+{\left(x-\sqrt{{x}^{2}-1}\right)}^{6}$

(iii)

(iv) ${\left(\sqrt{2}+1\right)}^{6}+{\left(\sqrt{2}-1\right)}^{6}$

(v) ${\left(3+\sqrt{2}\right)}^{5}-{\left(3-\sqrt{2}\right)}^{5}$

(vi) ${\left(2+\sqrt{3}\right)}^{7}+{\left(2-\sqrt{3}\right)}^{7}$

(vii) ${\left(\sqrt{3}+1\right)}^{5}-{\left(\sqrt{3}-1\right)}^{5}$

(viii) ${\left(0.99\right)}^{5}+{\left(1.01\right)}^{5}$

(ix) ${\left(\sqrt{3}+\sqrt{2}\right)}^{6}-{\left(\sqrt{3}-\sqrt{2}\right)}^{6}$

(x) ${\left\{{a}^{2}+\sqrt{{a}^{2}-1}\right\}}^{4}+{\left\{{a}^{2}-\sqrt{{a}^{2}-1}\right\}}^{4}$

(i)

(ii)
$\left(x+\sqrt{{x}^{2}-1}{\right)}^{6}+\left(x-\sqrt{{x}^{2}-1}{\right)}^{6}\phantom{\rule{0ex}{0ex}}=2\left[{}^{6}C_{0}{x}^{6}\left(\sqrt{{x}^{2}-1}{\right)}^{0}+{}^{6}C_{2}{x}^{4}\left(\sqrt{{x}^{2}-1}{\right)}^{2}+{}^{6}C_{4}{x}^{2}\left(\sqrt{{x}^{2}-1}{\right)}^{4}+{}^{6}C_{6}{x}^{0}\left(\sqrt{{x}^{2}-1}{\right)}^{6}\right]\phantom{\rule{0ex}{0ex}}=2\left[{x}^{6}+15{x}^{4}\left({x}^{2}-1\right)+15{x}^{2}\left({x}^{2}-1{\right)}^{2}+\left({x}^{2}-1{\right)}^{3}\right]\phantom{\rule{0ex}{0ex}}=2\left[{x}^{6}+15{x}^{6}-15{x}^{4}+15{x}^{2}\left({x}^{4}-2{x}^{2}+1\right)+\left({x}^{6}-1+3{x}^{2}-3{x}^{4}\right)\right]\phantom{\rule{0ex}{0ex}}=2\left[{x}^{6}+15{x}^{6}-15{x}^{4}+15{x}^{6}-30{x}^{4}+15{x}^{2}+{x}^{6}-1+3{x}^{2}-3{x}^{4}\right]\phantom{\rule{0ex}{0ex}}=64{x}^{6}-96{x}^{4}+36{x}^{2}-2$

(iii)
$\left(1+2\sqrt{x}{\right)}^{5}+\left(1-2\sqrt{x}{\right)}^{5}\phantom{\rule{0ex}{0ex}}=2\left[{}^{5}C_{0}\left(2\sqrt{x}{\right)}^{0}+{}^{5}C_{2}\left(2\sqrt{x}{\right)}^{2}+{}^{5}C_{4}\left(2\sqrt{x}{\right)}^{4}\right]\phantom{\rule{0ex}{0ex}}=2\left[1+10×4x+5×16{x}^{2}\right]\phantom{\rule{0ex}{0ex}}=2\left[1+40x+80{x}^{2}\right]$

(iv)

(v)
$\left(3+\sqrt{2}{\right)}^{5}-\left(3-\sqrt{2}{\right)}^{5}\phantom{\rule{0ex}{0ex}}=2\left[{}^{5}C_{1}×{3}^{4}×\left(\sqrt{2}{\right)}^{1}+{}^{5}C_{3}×{3}^{2}×\left(\sqrt{2}{\right)}^{3}+{}^{5}C_{5}×{3}^{0}×\left(\sqrt{2}{\right)}^{5}\right]$

$=2\left[5×81×\sqrt{2}+10×9×2\sqrt{2}+4\sqrt{2}\right]\phantom{\rule{0ex}{0ex}}=2\sqrt{2}\left(405+180+4\right)=1178\sqrt{2}$

(vi)
$\left(2+\sqrt{3}{\right)}^{7}+\left(2-\sqrt{3}{\right)}^{7}\phantom{\rule{0ex}{0ex}}=2\left[{}^{7}C_{0}×{2}^{7}×\left(\sqrt{3}{\right)}^{0}+{}^{7}C_{2}×{2}^{5}×\left(\sqrt{3}{\right)}^{2}+{}^{7}C_{4}×{2}^{3}×\left(\sqrt{3}{\right)}^{4}+{}^{7}C_{6}×{2}^{1}×\left(\sqrt{3}{\right)}^{6}\right]\phantom{\rule{0ex}{0ex}}=2\left[128+21×32×3+35×8×9+7×2×27\right]\phantom{\rule{0ex}{0ex}}=2\left[128+2016+2520+378\right]\phantom{\rule{0ex}{0ex}}=2×5042=10084$

(vii)
$\left(\sqrt{3}+1{\right)}^{5}-\left(\sqrt{3}-1{\right)}^{5}\phantom{\rule{0ex}{0ex}}=2\left[{}^{5}C_{1}×\left(\sqrt{3}{\right)}^{4}+{}^{5}C_{3}×\left(\sqrt{3}{\right)}^{2}+{}^{5}C_{5}×\left(\sqrt{3}{\right)}^{0}\right]\phantom{\rule{0ex}{0ex}}=2\left[5×9+10×3+1\right]\phantom{\rule{0ex}{0ex}}=2×76=152$

(viii)
$\left(0.99{\right)}^{5}+\left(1.01{\right)}^{5}\phantom{\rule{0ex}{0ex}}=\left(1-0.01{\right)}^{5}+\left(1+0.01{\right)}^{5}\phantom{\rule{0ex}{0ex}}=2\left[{}^{5}C_{0}\left(0.01{\right)}^{0}+{}^{5}C_{2}\left(0.01{\right)}^{2}+{}^{5}C_{4}\left(0.01{\right)}^{4}\right]\phantom{\rule{0ex}{0ex}}=2\left[1+10×0.0001+5×0.00000001\right]\phantom{\rule{0ex}{0ex}}=2×1.00100005=2.0020001$

(ix)
$\left(\sqrt{3}+\sqrt{2}{\right)}^{6}-\left(\sqrt{3}-\sqrt{2}{\right)}^{6}\phantom{\rule{0ex}{0ex}}=2\left[{}^{6}C_{1}\left(\sqrt{3}{\right)}^{5}\left(\sqrt{2}{\right)}^{1}+{}^{6}C_{3}\left(\sqrt{3}{\right)}^{3}\left(\sqrt{2}{\right)}^{3}+{}^{6}C_{5}\left(\sqrt{3}{\right)}^{1}\left(\sqrt{2}{\right)}^{5}\right]$
$=2\left[6×9\sqrt{3}×\sqrt{2}+20×3\sqrt{3}×2\sqrt{2}+6×\sqrt{3}×4\sqrt{2}\right]\phantom{\rule{0ex}{0ex}}=2\left[\sqrt{6}\left(54+120+24\right)\right]\phantom{\rule{0ex}{0ex}}=396\sqrt{6}\phantom{\rule{0ex}{0ex}}$

(x)
${\left\{{a}^{2}+\sqrt{{a}^{2}-1}\right\}}^{4}+{\left\{{a}^{2}-\sqrt{{a}^{2}-1}\right\}}^{4}\phantom{\rule{0ex}{0ex}}=2\left[{}^{4}C_{0}\left({a}^{2}{\right)}^{4}\left(\sqrt{{a}^{2}-1}{\right)}^{0}+{}^{4}C_{2}\left({a}^{2}{\right)}^{2}\left(\sqrt{{a}^{2}-1}{\right)}^{2}+{}^{4}C_{4}\left({a}^{2}{\right)}^{0}\left(\sqrt{{a}^{2}-1}{\right)}^{4}\right]\phantom{\rule{0ex}{0ex}}=2\left[{a}^{8}+6{a}^{4}\left({a}^{2}-1\right)+\left({a}^{2}-1{\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=2\left[{a}^{8}+6{a}^{6}-6{a}^{4}+{a}^{4}+1-2{a}^{2}\right]\phantom{\rule{0ex}{0ex}}=2{a}^{8}+12{a}^{6}-10{a}^{4}-4{a}^{2}+2$

Question 3:

Find ${\left(a+b\right)}^{4}-{\left(a-b\right)}^{4}$. Hence, evaluate ${\left(\sqrt{3}+\sqrt{2}\right)}^{4}-{\left(\sqrt{3}-\sqrt{2}\right)}^{4}$.

The expression $\left(a+b{\right)}^{4}-\left(a-b{\right)}^{4}$ can be written as

Question 4:

Find ${\left(x+1\right)}^{6}+{\left(x-1\right)}^{6}$. Hence, or otherwise evaluate ${\left(\sqrt{2}+1\right)}^{6}+\sqrt{2}-{1}^{6}$.

The expression ${\left(x+1\right)}^{6}+{\left(x-1\right)}^{6}$ can be written as
$\left(x+1{\right)}^{6}+\left(x-1{\right)}^{6}\phantom{\rule{0ex}{0ex}}=2\left[{}^{6}C_{0}{x}^{6}+{}^{6}C_{2}{x}^{4}+{}^{6}C_{4}{x}^{2}+{}^{6}C_{6}{x}^{0}\right]\phantom{\rule{0ex}{0ex}}=2\left[{x}^{6}+15{x}^{4}+15{x}^{2}+1\right]\phantom{\rule{0ex}{0ex}}$

By taking $x=\sqrt{2}$, we get:
$\left(\sqrt{2}+1{\right)}^{6}+\left(\sqrt{2}-1{\right)}^{6}=2\left[\left(\sqrt{2}{\right)}^{6}+15\left(\sqrt{2}{\right)}^{4}+15\left(\sqrt{2}{\right)}^{2}+1\right]\phantom{\rule{0ex}{0ex}}$
$\phantom{\rule{0ex}{0ex}}=2\left[8+15×4+15×2+1\right]\phantom{\rule{0ex}{0ex}}=2×\left(8+60+30+1\right)\phantom{\rule{0ex}{0ex}}=198$

Question 5:

Using binomial theorem evaluate each of the following:
(i) (96)3
(ii) (102)5
(iii) (101)4
(iv) (98)5

(i) (96)3
$=\left(100-4{\right)}^{3}\phantom{\rule{0ex}{0ex}}={}^{3}C_{0}×{100}^{3}×{4}^{0}-{}^{3}C_{1}×{100}^{2}×{4}^{1}+{}^{3}C_{2}×{100}^{1}×{4}^{2}-{}^{3}C_{3}×{100}^{0}×{4}^{3}\phantom{\rule{0ex}{0ex}}=1000000-120000+4800-64\phantom{\rule{0ex}{0ex}}=884736$

(ii) (102)5
$=\left(100+2{\right)}^{5}\phantom{\rule{0ex}{0ex}}={}^{5}C_{0}×{100}^{5}×{2}^{0}+{}^{5}C_{1}×{100}^{4}×{2}^{1}+{}^{5}C_{2}×{100}^{3}×{2}^{2}+{}^{5}C_{3}×{100}^{2}×{2}^{3}+{}^{5}C_{4}×{100}^{1}×{2}^{4}+{}^{5}C_{5}×{100}^{0}×{2}^{5}\phantom{\rule{0ex}{0ex}}=10000000000+1000000000+40000000+800000+8000+32\phantom{\rule{0ex}{0ex}}=11040808032\phantom{\rule{0ex}{0ex}}$

(iii) (101)4
$=\left(100+1{\right)}^{4}\phantom{\rule{0ex}{0ex}}={}^{4}C_{0}×{100}^{4}+{}^{4}C_{1}×{100}^{3}+{}^{4}C_{2}×{100}^{2}+{}^{4}C_{3}×{100}^{1}+{}^{4}C_{4}×{100}^{0}\phantom{\rule{0ex}{0ex}}=100000000+4000000+60000+400+1\phantom{\rule{0ex}{0ex}}=104060401\phantom{\rule{0ex}{0ex}}$

(iv) (98)5
$\left(100-2{\right)}^{5}\phantom{\rule{0ex}{0ex}}={}^{5}C_{0}×{100}^{5}×{2}^{0}+{-}^{5}{C}_{1}×{100}^{4}×{2}^{1}+{}^{5}C_{2}×{100}^{3}×{2}^{2}-{}^{5}C_{3}×{100}^{2}×{2}^{3}+{}^{5}C_{4}×{100}^{1}×{2}^{4}-{}^{5}C_{5}×{100}^{0}×{2}^{5}\phantom{\rule{0ex}{0ex}}=10000000000-1000000000+40000000-800000+8000-32\phantom{\rule{0ex}{0ex}}=9039207968$

Question 6:

Using binomial theorem, prove that ${2}^{3n}-7n-1$ is divisible by 49, where .

${2}^{3n}-7n-1={8}^{n}-7n-1$                ...(1)

Question 7:

Using binomial theorem, prove that ${3}^{2n+2}-8n-9$ is divisible by 64, .

Consider

Question 8:

If n is a positive integer, prove that ${3}^{3n}-26n-1$ is divisible by 676.

Question 9:

Using binomial theorem, indicate which is larger (1.1)10000 or 1000.

We have:
(1.1)10000

Question 10:

Using binomial theorem determine which number is larger (1.2)4000 or 800?

We have:
(1.2)4000$=\left(1+0.2{\right)}^{4000}\phantom{\rule{0ex}{0ex}}={}^{4000}C_{0}+{}^{4000}C_{1}×\left(0.2{\right)}^{1}+{}^{4000}C_{2}×\left(0.2{\right)}^{2}+...{}^{4000}C_{4000}×\left(0.2{\right)}^{4000}$

Hence, (1.2)4000 is greater than 800

Question 11:

Find the value of (1.01)10 + (1 − 0.01)10 correct to 7 places of decimal.

Hence, the value of (1.01)10 + (1 − 0.01)10 correct to 7 places of the decimal is 2.0090042

Question 12:

Show that ${2}^{4n+4}-15n-16$, where n ∈ $\mathrm{ℕ}$ is divisible by 225.

We have,

Thus, ​ ${2}^{4n+4}-15n-16$, where n ∈ $\mathrm{ℕ}$ is divisible by 225.

Question 1:

Find the 11th term from the beginning and the 11th term from the end in the expansion of ${\left(2x-\frac{1}{{x}^{2}}\right)}^{25}$.

Given:
${\left(2x-\frac{1}{{x}^{2}}\right)}^{25}$
Clearly, the given expression contains 26 terms.

So, the 11th term from the end is the (26 − 11 + 1)th term from the beginning. In other words, the 11th term from the end is the 16th term from the beginning.

Thus, we have:

Now, we will find the 11th term from the beginning.

Question 2:

Find the 7th term in the expansion of ${\left(3{x}^{2}-\frac{1}{{x}^{3}}\right)}^{10}$.

We need to find the 7th term of the given expression.
Let it be T7
Now, we have
${T}_{7}={T}_{6+1}$
$={}^{10}C_{6}\left(3{x}^{2}{\right)}^{10-6}{\left(\frac{-1}{{x}^{3}}\right)}^{6}\phantom{\rule{0ex}{0ex}}={}^{10}C_{6}\left({3}^{4}\right)\left({x}^{8}\right)\left(\frac{1}{{x}^{18}}\right)\phantom{\rule{0ex}{0ex}}=\frac{10×9×8×7×81}{4×3×2×{x}^{10}}=\frac{17010}{{x}^{10}}$

Thus, the 7th term of the given expression is $\frac{17010}{{x}^{10}}$

Question 3:

Find the 5th term from the end in the expansion of ${\left(3x-\frac{1}{{x}^{2}}\right)}^{10}$

Given:
${\left(3x-\frac{1}{{x}^{2}}\right)}^{10}$
Clearly, the expression has 6 terms.
The 5th term from the end is the (11 − 5 + 1)th, i.e., 7th, term from the beginning.
Thus, we have:

${T}_{7}={T}_{6+1}\phantom{\rule{0ex}{0ex}}={}^{10}C_{6}\left(3x{\right)}^{10-6}{\left(\frac{-1}{{x}^{2}}\right)}^{6}\phantom{\rule{0ex}{0ex}}={}^{10}C_{6}\left({3}^{4}\right)\left({x}^{4}\right)\left(\frac{1}{{x}^{12}}\right)\phantom{\rule{0ex}{0ex}}=\frac{10×9×8×7×81}{4×3×2×1×{x}^{8}}=\frac{17010}{{x}^{8}}$

Question 4:

Find the 8th term in the expansion of .

We need to find the 8th term in the given expression.
$\because {T}_{8}={T}_{7+1}$

Question 5:

Find the 7th term in the expansion of ${\left(\frac{4x}{5}+\frac{5}{2x}\right)}^{8}$.

We need to find the 7th term in the given expression.

$\because {T}_{7}={T}_{6+1}$

Question 6:

Find the 4th term from the beginning and 4th term from the end in the expansion of ${\left(x+\frac{2}{x}\right)}^{9}$.

Let Tr+1 be the 4th term from the end.
Then, Tr+1 is (10 − 4 + 1)th, i.e., 7th, term from the beginning.

4th term from the beginning = ${T}_{4}={T}_{3+1}$

Question 7:

Find the 4th term from the end in the expansion of ${\left(\frac{4x}{5}-\frac{5}{2x}\right)}^{8}$.

Let Tr+1 be the 4th term from the end of the given expression.
Then,
Tr+1 is (10 − 4 + 1)th term, i.e., 7th term, from the beginning.
Thus, we have:

Question 8:

Find the 7th term from the end in the expansion of ${\left(2{x}^{2}-\frac{3}{2x}\right)}^{8}$

Let Tr+1 be the 7th term from the end in the given expression.
Then, we have:
Tr+1 = (9 − 7 + 1) =  3rd term from the beginning
Now,

Question 9:

Find the coefficient of:
(i) x10 in the expansion of ${\left(2{x}^{2}-\frac{1}{x}\right)}^{20}$

(ii) x7 in the expansion of ${\left(x-\frac{1}{{x}^{2}}\right)}^{40}$

(iii) ${x}^{-15}$ in the expansion of ${\left(3{x}^{2}-\frac{a}{3{x}^{3}}\right)}^{10}$

(iv) ${x}^{9}$ in the expansion of ${\left({x}^{2}-\frac{1}{3x}\right)}^{9}$

(v) ${x}^{m}$ in the expansion of ${\left(x+\frac{1}{x}\right)}^{n}$

(vi) x in the expansion of .

(vii) ${a}^{5}{b}^{7}$ in the expansion of ${\left(a-2b\right)}^{12}$.

(viii) x in the expansion of .

(i) Suppose x10 occurs in the (+ 1)th term in the given expression.

Then, we have:

Here,

(ii) Suppose x7 occurs at the (+ 1) th term in the given expression.

Then, we have:

(iii)  Suppose x−15 occurs at the (+ 1)th term in the given expression.
Then, we have:

(iv) Suppose x9 occurs at the (+ 1)th term in the above expression.

Then, we have:

(v)
Suppose xm occurs at the (+ 1)th term in the given expression.

Then, we have:

(vi) Suppose x occurs at the (+ 1)th term in the given expression.

Then, we have:

(vii)
Suppose a5 b7 occurs at the (r + 1)th term in the given expression.

Then, we have:

(viii) Suppose x occurs at the (+ 1)th term in the given expression.

Then, we have:

Question 10:

Which term in the expansion of ${\left\{{\left(\frac{x}{\sqrt{y}}\right)}^{1/3}+{\left(\frac{y}{{x}^{1/3}}\right)}^{1/2}\right\}}^{21}$ contains x and y to one and the same power?

Suppose Tr+1th term in the given expression contains x and y to one and the same power.
Then,

Question 11:

Does the expansion of $\left(2{x}^{2}-\frac{1}{x}\right)$ contain any term involving x9?

Suppose x9 occurs in the given expression at the (r + 1)th term.
Then, we have:

Hence, there is no term with x9 in the given expression.

Question 12:

Show that the expansion of ${\left({x}^{2}+\frac{1}{x}\right)}^{12}$ does not contain any term involving x−1.

Suppose x1 occurs at the (r + 1)th term in the given expression.
Then,

Hence, the expansion of ${\left({x}^{2}+\frac{1}{x}\right)}^{12}$ does not contain any term involving x−1.

Question 13:

Find the middle term in the expansion of:
(i) ${\left(\frac{2}{3}x-\frac{3}{2x}\right)}^{20}$

(ii) ${\left(\frac{a}{x}+bx\right)}^{12}$

(iii) ${\left({x}^{2}-\frac{2}{x}\right)}^{10}$

(iv) ${\left(\frac{x}{a}-\frac{a}{x}\right)}^{10}$

(i) Here,
n = 20  (Even number)
Therefore, the middle term is the $\left(\frac{n}{2}+1\right)$th term, i.e., the 11th term.

(ii) Here,
n = 12 (Even number)
Therefore, the middle term is the $\left(\frac{n}{2}+1\right)\mathrm{th}$  i.e. 7th term

(iii) Here,
n = 10    (Even number)
Therefore, the middle term is the $\left(\frac{n}{2}+1\right)\mathrm{th}$  i.e. 6th term

(iv) Here,
n = 10 (Even number)
Therefore, the middle term is the $\left(\frac{n}{2}+1\right)\mathrm{th}$  i.e. 6th term

Question 14:

Find the middle terms in the expansion of:
(i) ${\left(3x-\frac{{x}^{3}}{6}\right)}^{9}$

(ii) ${\left(2{x}^{2}-\frac{1}{x}\right)}^{7}$

(iii) ${\left(3x-\frac{2}{{x}^{2}}\right)}^{15}$

(iv) ${\left({x}^{4}-\frac{1}{{x}^{3}}\right)}^{11}$

(i) Here, n, i.e. 9, is an odd number.
Thus, the middle terms are

(ii) Here, n, i.e., 7, is an odd number.

(iii)

(iv)

Question 15:

Find the middle terms(s) in the expansion of:
(i) ${\left(x-\frac{1}{x}\right)}^{10}$

(ii) ${\left(1-2x+{x}^{2}\right)}^{n}$

(iii) ${\left(1+3x+3{x}^{2}+{x}^{3}\right)}^{2n}$

(iv) ${\left(2x-\frac{{x}^{2}}{4}\right)}^{9}$

(v) ${\left(x-\frac{1}{x}\right)}^{2n+1}$

(vi) ${\left(\frac{x}{3}+9y\right)}^{10}$

(vii) ${\left(3-\frac{{x}^{3}}{6}\right)}^{7}$

(viii) ${\left(2ax-\frac{b}{{x}^{2}}\right)}^{12}$

(ix) ${\left(\frac{p}{x}+\frac{x}{p}\right)}^{9}$

(x) ${\left(\frac{x}{a}-\frac{a}{x}\right)}^{10}$

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

Question 16:

Find the term independent of x in the expansion of the following expressions:
(i) ${\left(\frac{3}{2}{x}^{2}-\frac{1}{3x}\right)}^{9}$

(ii) ${\left(2x+\frac{1}{3{x}^{2}}\right)}^{9}$

(iii) ${\left(2{x}^{2}-\frac{3}{{x}^{3}}\right)}^{25}$

(iv) ${\left(3x-\frac{2}{{x}^{2}}\right)}^{15}$

(v) ${\left(\frac{\sqrt{x}}{3}+\frac{3}{2{x}^{2}}\right)}^{10}$

(vi) ${\left(x-\frac{1}{{x}^{2}}\right)}^{3n}$

(vii) ${\left(\frac{1}{2}{x}^{1/3}+{x}^{-1/5}\right)}^{8}$

(viii) $\left(1+x+2{x}^{3}\right){\left(\frac{3}{2}{x}^{2}-\frac{3}{3x}\right)}^{9}$

(ix)

(x) ${\left(\frac{3}{2}{x}^{2}-\frac{1}{3x}\right)}^{6}$

(i) Suppose the (r + 1)th term in the given expression is independent of x.
Now,

(ii) Suppose the (r + 1)th term in the given expression is independent of x.
Now,

(iii) Suppose the (r + 1)th term in the given expression is independent of x.
Now,

(iv)  Suppose the (r + 1)th term in the given expression is independent of x.
Now,

(v) Suppose the (r + 1)th term in the given expression is independent of x.
Now,

(vi)  Suppose the (r + 1)th term in the given expression is independent of x.
Now,

(vii)  Suppose the (r + 1)th term in the given expression is independent of x.
Now,

(ix) Suppose the (r + 1)th term in the given expression is independent of x.
Now,

(x) Suppose the (r + 1)th term in the given expression is independent of x.
Now,

Question 17:

If the coefficients of th terms in the expansion of ${\left(1+x\right)}^{18}$ are equal, find r.

Question 18:

If the coefficients of (2r + 1)th term and (r + 2)th term in the expansion of (1 + x)43 are equal, find r.

Question 19:

Prove that the coefficient of (r + 1)th term in the expansion of (1 + x)n + 1 is equal to the sum of the coefficients of rth and (r + 1)th terms in the expansion of (1 + x)n.

Question 20:

Prove that the term independent of x in the expansion of ${\left(x+\frac{1}{x}\right)}^{2n}$ is

Question 21:

The coefficients of 5th, 6th and 7th terms in the expansion of (1 + x)n are in A.P., find n.

Question 22:

If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n are in A.P., show that $2{n}^{2}-9n+7=0$.

Hence proved.

Question 23:

If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)n are in A.P., then find the value of n.

Coefficients of the 2nd, 3rd  and 4th terms in the given expansion are:

Question 24:

If in the expansion of (1 + x)n, the coefficients of pth and qth terms are equal, prove that p + q = n + 2, where $p\ne q$.

If $p\ne q$, then $p+q=n+2$
Hence proved

Question 25:

Find a, if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

We have Coefficient of x2 Coefficient of x3

Question 26:

Find the coefficient of a4 in the product (1 + 2a)4 (2 − a)5 using binomial theorem.

Question 27:

In the expansion of (1 + x)n the binomial coefficients of three consecutive terms are respectively 220, 495 and 792, find the value of n.

Question 28:

If in the expansion of (1 + x)n, the coefficients of three consecutive terms are 56, 70 and 56, then find n and the position of the terms of these coefficients.

Question 29:

If 3rd, 4th 5th and 6th terms in the expansion of (x + a)n be respectively a, b, c and d, prove that $\frac{{b}^{2}-ac}{{c}^{2}-bd}=\frac{5a}{3c}$.

Question 30:

If a, b, c and d in any binomial expansion be the 6th, 7th, 8th and 9th terms respectively, then prove that $\frac{{b}^{2}-ac}{{c}^{2}-bd}=\frac{4a}{3c}$.

Question 31:

If the coefficients of three consecutive terms in the expansion of (1 + x)n be 76, 95 and 76, find n.

Question 32:

If the 6th, 7th and 8th terms in the expansion of (x + a)n are respectively 112, 7 and 1/4, find x, a, n.

According to the question,

Question 33:

If the 2nd, 3rd and 4th terms in the expansion of (x + a)n are 240, 720 and 1080 respectively, find x, a, n.

Question 34:

Find a, b and n in the expansion of (a + b)n, if the first three terms in the expansion are 729, 7290 and 30375 respectively.

Question 35:

If the term free from x in the expansion of ${\left(\sqrt{x}-\frac{k}{{x}^{2}}\right)}^{10}$ is 405, find the value of k.

Let (r + 1)th term, in the expansion of ${\left(\sqrt{x}-\frac{k}{{x}^{2}}\right)}^{10}$, be free from x and be equal to Tr + 1. Then,

If Tr + 1 is independent of x, then
$5-\frac{5r}{2}=0⇒r=2$

Putting r = 2 in (1), we obtain

But it is given that the value of the term free from x is 405.
$\therefore 45{k}^{2}=405⇒{k}^{2}=9⇒k=±3$

Hence, the value of k is $±3$.

Question 36:

Find the sixth term in the expansion ${\left({y}^{\frac{1}{2}}+{x}^{\frac{1}{3}}\right)}^{n}$, if the binomial coefficient of the third term from the end is 45.

In the binomial expansion of ${\left({y}^{\frac{1}{2}}+{x}^{\frac{1}{3}}\right)}^{n}$, there are (n + 1) terms.

The third term from the end in the expansion of ${\left({y}^{\frac{1}{2}}+{x}^{\frac{1}{3}}\right)}^{n}$, is the third term from the beginning in the expansion of ${\left({x}^{\frac{1}{3}}+{y}^{\frac{1}{2}}\right)}^{n}$.

∴ The binomial coefficient of the third term from the end = ${}^{n}{C}_{2}$

If is given that the binomial coefficient of the third term from the end is 45.

Let Tbe the sixth term in the binomial expansion of ${\left({y}^{\frac{1}{2}}+{x}^{\frac{1}{3}}\right)}^{n}$. Then

Hence, the sixth term in the expansion of ${\left({y}^{\frac{1}{2}}+{x}^{\frac{1}{3}}\right)}^{n}$, is .

Question 37:

If p is a real number and if the middle term in the expansion of ${\left(\frac{p}{2}+2\right)}^{8}$ is 1120, find p.

In the binomial expansion of ${\left(\frac{p}{2}+2\right)}^{8}$, we observe that ${\left(\frac{8}{2}+1\right)}^{\mathrm{th}}$ i.e., 5th term is the middle term.

It is given that the middle term is 1120.

Hence, the real values of p is $±2$.

Question 38:

Find n in the binomial ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^{n}$, if the ratio of 7th term from the beginning to the 7th term from the end is $\frac{1}{6}$.

In the binomail expansion of ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^{n}$${\left[\left(n+1\right)-7+1\right]}^{\mathrm{th}}$ i.e., (n − 5)th term from the beginning is the 7th term from the end.

Now,
${T}_{7}{=}^{n}{C}_{6}{\left(\sqrt[3]{2}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^{6}{=}^{n}{C}_{6}×{2}^{\frac{n}{3}-2}×\frac{1}{{3}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}{T}_{n-5}{=}^{n}{C}_{n-6}{\left(\sqrt[3]{2}\right)}^{6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n-6}{=}^{n}{C}_{6}×{2}^{2}×\frac{1}{{3}^{\frac{n}{3}-2}}$

It is given that,
$\frac{{T}_{7}}{{T}_{n-5}}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{{}^{n}{C}_{6}×{2}^{\frac{n}{3}-2}×\frac{1}{{3}^{2}}}{{}^{n}{C}_{6}×{2}^{2}×\frac{1}{{3}^{\frac{n}{3}-2}}}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒{2}^{\frac{n}{3}-2-2}×{3}^{\frac{n}{3}-2-2}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{1}{6}\right)}^{4-\frac{n}{3}}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒4-\frac{n}{3}=1\phantom{\rule{0ex}{0ex}}⇒n=9$

Hence, the value of is 9.

Question 39:

if the seventh term from the beginning and end in the binomial expansion of ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^{n}$ are equal, find n.

In the binomail expansion of ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^{n}$${\left[\left(n+1\right)-7+1\right]}^{\mathrm{th}}$ i.e., (n − 5)th term from the beginning is the 7th term from the end.

Now,
${T}_{7}{=}^{n}{C}_{6}{\left(\sqrt[3]{2}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^{6}{=}^{n}{C}_{6}×{2}^{\frac{n}{3}-2}×\frac{1}{{3}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}{T}_{n-5}{=}^{n}{C}_{n-6}{\left(\sqrt[3]{2}\right)}^{6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n-6}{=}^{n}{C}_{6}×{2}^{2}×\frac{1}{{3}^{\frac{n}{3}-2}}$

It is given that,
${T}_{7}={T}_{n-5}\phantom{\rule{0ex}{0ex}}{⇒}^{n}{C}_{6}×{2}^{\frac{n}{3}-2}×\frac{1}{{3}^{2}}{=}^{n}{C}_{6}×{2}^{2}×\frac{1}{{3}^{\frac{n}{3}-2}}\phantom{\rule{0ex}{0ex}}⇒\frac{{2}^{\frac{n}{3}-2}}{{2}^{2}}=\frac{{3}^{2}}{{3}^{\frac{n}{3}-2}}\phantom{\rule{0ex}{0ex}}⇒{\left(6\right)}^{\frac{n}{3}-2}={6}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{n}{3}-2=2\phantom{\rule{0ex}{0ex}}⇒n=12$

Hence, the value of is 12.

Question 1:

If in the expansion of (1 + x)20, the coefficients of rth and (r + 4)th terms are equal, then r is equal to
(a) 7
(b) 8
(c) 9
(d) 10

(c) 9

Question 2:

The term without x in the expansion of ${\left(2x-\frac{1}{2{x}^{2}}\right)}^{12}$ is
(a) 495
(b) −495
(c) −7920
(d) 7920

(d) 7920

Question 3:

If rth term in the expansion of ${\left(2{x}^{2}-\frac{1}{x}\right)}^{12}$ is without x, then r is equal to
(a) 8
(b) 7
(c) 9
(d) 10

(c) 9

Question 4:

If in the expansion of (a + b)n and (a + b)n + 3, the ratio of the coefficients of second and third terms, and third and fourth terms respectively are equal, then n is
(a) 3
(b) 4
(c) 5
(d) 6

(c) n = 5

Question 5:

If A and B are the sums of odd and even terms respectively in the expansion of (x + a)n, then (x + a)2n − (xa)2n is equal to
(a) 4 (A + B)
(b) 4 (AB)
(c) AB
(d) 4 AB

(d) 4AB

Question 6:

The number of irrational terms in the expansion of ${\left({4}^{1/5}+{7}^{1/10}\right)}^{45}$ is
(a) 40
(b) 5
(c) 41
(d) none of these

(c) 41

Question 7:

The coefficient of ${x}^{-17}$ in the expansion of ${\left({x}^{4}-\frac{1}{{x}^{3}}\right)}^{15}$ is
(a) 1365
(b) −1365
(c) 3003
(d) −3003

(b) −1365

Question 8:

In the expansion of ${\left({x}^{2}-\frac{1}{3x}\right)}^{9}$, the term without x is equal to
(a) $\frac{28}{81}$

(b) $\frac{-28}{243}$

(c) $\frac{28}{243}$

(d) none of these

(c) $\frac{28}{243}$

Suppose the (r + 1)th term in the given expansion is independent of x.
Then , we have:

Question 9:

If an the expansion of ${\left(1+x\right)}^{15}$, the coefficients of terms are equal, then the value of r is
(a) 5
(b) 6
(c) 4
(d) 3

(a) 5

Question 10:

The middle term in the expansion of ${\left(\frac{2{x}^{2}}{3}+\frac{3}{2{x}^{2}}\right)}^{10}$ is
(a) 251
(b) 252
(c) 250
(d) none of these

(b) 252

Question 11:

If in the expansion of ${\left({x}^{4}-\frac{1}{{x}^{3}}\right)}^{15}$, ${x}^{-17}$ occurs in rth term, then
(a) r = 10
(b) r = 11
(c) r = 12
(d) r = 13

(c) r = 12

Here,

Question 12:

In the expansion of ${\left(x-\frac{1}{3{x}^{2}}\right)}^{9}$, the term independent of x is
(a) T3
(b) T4
(c) T5
(d) none of these

(b) T4

Question 13:

If in the expansion of (1 + y)n, the coefficients of 5th, 6th and 7th terms are in A.P., then n is equal to
(a) 7, 11
(b) 7, 14
(c) 8, 16
(d) none of these

(b) 7, 14

Question 14:

In the expansion of ${\left(\frac{1}{2}{x}^{1/3}+{x}^{-1/5}\right)}^{8}$, the term independent of x is
(a) T5
(b) T6
(c) T7
(d) T8

(b) T6
Suppose the (r + 1)th term in the given expansion is independent of x.
Thus, we have:

Question 15:

If the sum of odd numbered terms and the sum of even numbered terms in the expansion of ${\left(x+a\right)}^{n}$ are A and B respectively, then the value of ${\left({x}^{2}-{a}^{2}\right)}^{n}$ is
(a) ${A}^{2}-{B}^{2}$
(b) ${A}^{2}+{B}^{2}$
(c) 4 AB
(d) none of these

(a) ${A}^{2}-{B}^{2}$

Question 16:

If the coefficient of x in ${\left({x}^{2}+\frac{\lambda }{x}\right)}^{5}$ is 270, then $\lambda =$
(a) 3
(b) 4
(c) 5
(d) none of these

(a) 3

Question 17:

The coefficient of x4 in ${\left(\frac{x}{2}-\frac{3}{{x}^{2}}\right)}^{10}$ is
(a) $\frac{405}{256}$

(b) $\frac{504}{259}$

(c) $\frac{450}{263}$

(d) none of these

(a) $\frac{405}{256}$

Question 18:

The total number of terms in the expansion of ${\left(x+a\right)}^{100}+{\left(x-a\right)}^{100}$ after simplification is
(a) 202
(b) 51
(c) 50
(d) none of these

(b) 51
Here, n, i.e., 100, is even.
∴ Total number of terms in the expansion =

Question 19:

If ${T}_{2}/{T}_{3}$ in the expansion of in the expansion of ${\left(a+b\right)}^{n+3}$ are equal, then n =
(a) 3
(b) 4
(c) 5
(d) 6

(c) 5

Question 20:

The coefficient of $\frac{1}{x}$ in the expansion of is
(a)

(b)

(c)

(d) none of these

(b)

Question 21:

If the sum of the binomial coefficients of the expansion ${\left(2x+\frac{1}{x}\right)}^{n}$ is equal to 256, then the term independent of x is
(a) 1120
(b) 1020
(c) 512
(d) none of these

(a) 1120

Question 22:

If the fifth term of the expansion does not contain 'a'. Then n is equal to
(a) 2
(b) 5
(c) 10
(d) none of these

(c) 10

Question 23:

The coefficient of ${x}^{-3}$ in the expansion of ${\left(x-\frac{m}{x}\right)}^{11}$ is
(a)
(b)
(c)
(d)

(d)

Question 24:

The coefficient of the term independent of x in the expansion of ${\left(ax+\frac{b}{x}\right)}^{14}$ is
(a)

(b)

(c)

(d) $\frac{14!}{{\left(7!\right)}^{3}}{a}^{7}{b}^{7}$

(c)

Question 25:

The coefficient of x5 in the expansion of
(a) 51C5
(b) 9C5
(c) 31C621C6
(d) 30C5 + 20C5

(c) 31C621C6

Question 26:

The coefficient of x8 y10 in the expansion of (x + y)18 is
(a) 18C8
(b) 18p10
(c) 218
(d) none of these

(a) 18C8

Question 27:

If the coefficients of the (n + 1)th term and the (n + 3)th term in the expansion of (1 + x)20 are equal, then the value of n is
(a) 10
(b) 8
(c) 9
(d) none of these

(c) 9

Question 28:

If the coefficients of 2nd, 3rd and 4th terms in the expansion of are in A.P., then n =
(a) 7
(b) 14
(c) 2
(d) none of these

(a) 7
Coefficients of the 2nd, 3rd  and 4th terms in the given expansion are:

Question 29:

The middle term in the expansion of ${\left(\frac{2x}{3}-\frac{3}{2{x}^{2}}\right)}^{2n}$ is
(a) ${}^{2n}C_{n}$

(b)

(c) ${}^{2n}C_{n}{x}^{-n}$

(d) none of these

(b)

Question 30:

If rth term is the middle term in the expansion of ${\left({x}^{2}-\frac{1}{2x}\right)}^{20}$, then ${\left(r+3\right)}^{th}$ term is
(a) ${}^{20}C_{14}\left(\frac{x}{{2}^{14}}\right)$

(b)

(c)

(d) none of these

(c)
Here n is even
So, The middle term in the given expansion is
Therefore, (r + 3)th term is the 14th term.

Question 31:

The number of terms with integral coefficients in the expansion of is
(a) 100
(b) 50
(c) 150
(d) 101

(d) 101

Question 32:

Constant term in the expansion of ${\left(x-\frac{1}{x}\right)}^{10}$ is
(a) 152
(b) −152
(c) −252
(d) 252

(c) −252

Suppose (r + 1)th term is the constant term in the given expansion.
Then, we have:

Question 33:

If the coefficients of x2 and x3 in the expansion of (3 + ax)9 are the same, then the value of a is
(a) $-\frac{7}{9}$

(b) $-\frac{9}{7}$

(c) $\frac{7}{9}$

(d) $\frac{9}{7}$

(d) $\frac{9}{7}$
Coefficients of x2 Coefficients of x3

Question 34:

Given the integers r > 1, n > 2, and coefficient of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then
(a) n = 2r
(b) n = 3r
(c) n = 2r + 1
(d) none of these

Given r > 1 and n > 2
and coefficient of T3r = coefficient of Tr+2 is expansion of (1 + x)2n

Hence, the correct answer is option A.

Question 35:

The two successive terms in the expansion of (1 + x)24 whose coefficients are in the ratio 1 : 4 are
(a) 3rd and 4th
(b) 4th and 5th
(c) 5th and 6th
(d) 6th and 7th

For (1 + x)24 two successive terms have coefficients in ration 1 : 4
Let the two successive terms be (r + 1)th and (r + 2) terms

Hence 5th  and 6th terms.
Hence, the correct answer is option C.

Question 36:

If the coefficients of 2nd, 3rd and the 4th terms in the expansion of (1 + x)n are in A.P., then the value of n is
(a) 2
(b) 7
(c) 11
(d) 14

Given coefficient of 2nd, 3rd and 4th terms in the expansion (1 + x)n are A.P
Since coefficient of (r + 1)th terms is nCr
∴ 2nd, 3rd and 4th coefficient are such that 2nCr = nC1 + nC3

Hence, the correct answer is option B.

Question 37:

If the middle term of is equal to $7\frac{7}{8},$ then the value of x is
(a) $2n\pi +\frac{\mathrm{\pi }}{6}$

(b) $n\pi +\frac{\mathrm{\pi }}{6}$

(c) $n\pi +{\left(-1\right)}^{n}\frac{\mathrm{\pi }}{6}$

(d) $n\pi +{\left(-1\right)}^{n}\frac{\mathrm{\pi }}{3}$

Given middle term of
Since n = 10
i.e. middle term is ${\left(\frac{n}{2}+1\right)}^{\mathrm{th}}$ term is 6th term.

Hence, the correct answer is option C.

Question 38:

The total number of terms in the expansion of (x + a)51 – (xa)51 after simplification is
(a) 102
(b) 25
(c) 26
(d) none of these

i.e.      1st, 3rd, 5th, 7th ________  49th, 51th term are there
applying A.P.
a + ( n – 1)d = 51
i.e(n – 1) 51
i.e2(n – 150
i.e26

Hencethe correct answer is option C.

Question 39:

If the coefficients of x7 and x8 in ${\left(2+\frac{x}{3}\right)}^{n}$ are equal, then n is
(a) 56
(b) 55
(c) 45
(d) 15

Coefficient of x7 is

and coefficient of x8 is

Since T8 = T9 i.e coefficient of T8 = coefficient of T9

Hence, the correct answer is option B.

Question 40:

The ratio of the coefficient of x15 to the term independent of x in ${\left({x}^{2}+\frac{2}{x}\right)}^{15},$ is
(a) 12 : 32
(b) 1 : 32
(c) 32 : 12
(d) 32 : 1

In we have ${T}_{r+1}{=}^{15}{C}_{r}{\left({x}^{2}\right)}^{15-r}{\left(\frac{2}{x}\right)}^{r}$

Hence for the term independent of x,
30 – 3r = 0
i.e. r = 10
hence T11 has coefficient 15C10 210       ...(1)
and term with x15 will have 30 – 3r = 15
i.e. 15 = 3r
i.e. r = 5
∴ coefficient will be 15C5 25              ...(2)
∴  ratio of coefficient of x15 to the term independent of x will be

i.e. ratio will be 1 : 32
Hence, the correct answer is option B.

Question 41:

If $z={\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)}^{5}+{\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)}^{5},$ then
(a) Re (z) = 0
(b) Im (z) = 0
(c) Re (z) > 0, Im (z) > 0
(d) Re (z) > 0, Im (z) < 0

Question 42:

If (1 – x + x2)n = a0 + a1 x + a2 x2 +...+a2n  x2n, then a0 + a2 + a4 +...+ a2n equals

(a) $\frac{{3}^{n}+1}{2}$

(b) $\frac{{3}^{n}-1}{2}$

(c) $\frac{1-{3}^{n}}{2}$

(d) ${3}^{n}+\frac{1}{2}$

Given:

i.e. all even terms are involved
∴ replace x by 1 in equation (1)
we get

and now replace x by –1 in equation (1), we get

By adding (2) and (3), we get

Hence, the correct answer is option A.

Question 1:

The largest coefficient in (1 + x)30 is ___________.

i.e for middle term
which is 30C15.

Question 2:

The largest coefficient in (1 + x)41 is ___________.

In (1 + x)41
Since n is odd i.e 41
∴ The largest coefficient is 41C21 or 41C20
term gives largest coefficient

Question 3:

The number of terms in the expansion of (x + y + z)n is ___________.

∴ Number of terms in the expansion

Question 4:

Middle term in the expansion of (a3 + ba)28 is ___________.

In (a3 + ba)28
Here n = 28
So, there is one middle term

Question 5:

The ratio of the coefficients of xm and xn in the expansion of (1 + x)m + n is ___________.

For (1 + x)m + n
Tr+1 = m+nCr xr
for coefficient of xm, r = m
i.e coefficient is m+nCm
and for coefficient of xn, r = n
i.e. coefficient is m+nCn

Question 6:

The coefficient of a–6 b4 in the expansion ${\left(\frac{1}{a}-\frac{2}{3}b\right)}^{10}$ is ___________.

We get

Question 7:

In the expansion of ${\left({x}^{2}-\frac{1}{{x}^{2}}\right)}^{16},$ the value of the constant term is ___________.

i.e value of constant term is 16C8.

Question 8:

The position of the term independent of x in the expansion of ${\left(\sqrt{\frac{x}{3}}+\frac{3}{2{x}^{2}}\right)}^{10}$ is ___________.

Hence, third term is independent of x.

Question 9:

If 215 is divided by 13, the remainder is ___________.

If 215 is divided by 13
Since 215 = (5 – 3)15
= 10C0 (5)15 (–3)0 + ..........
Correction

Question 10:

The sum of the series $\sum _{r=0}^{10}$ 20Cr is ___________.

Question 11:

The number of terms in the expansion of ${\left\{{\left(2x+{y}^{3}\right)}^{4}\right\}}^{7}$ is ___________.

In {(2x + y3)4}7
n = ??
Since In (a + b)n ; number of terms is n + 1
∴ {(2x + y3)}28
Number of terms is 28 + 1 = 29

Question 12:

The middle term in the expansion of ${\left(x-\frac{1}{x}\right)}^{18}$ is ___________.

${\left(x-\frac{1}{x}\right)}^{18}$
Since 18 is even
there is only one middle term i.e. ${\left(\frac{2n}{2}+1\right)}^{\mathrm{th}}$term
i.e. (n + 1)th term

Question 13:

The coefficient of the middle term in the expansion of (1 + x)10 is ___________.

In (1 + x)10
Tr +1 = 10Cr xr
Middle term is obtained when r = 5
i.e. Tr +1 = 10C5 x5
i.e. coefficient of middle term is 10C5

Question 14:

The total number of terms in the expansion of (1 + x)2n – (1 – x)2n ___________.

Subtracting above two,

Question 15:

If x4 occurs in the rth terms in the expansion of ${\left({x}^{4}+\frac{1}{{x}^{3}}\right)}^{15},$ then r = ___________.

n = 15

To get x4, 7r – 45 = 4
i.e. 7r = 49
i.e r = 7

Question 16:

The coefficient of x in the binomial expansion of ${\left({x}^{2}+\frac{a}{x}\right)}^{5}$ is ___________.

For coefficient  of x, put 3r – 5 = 1
i.e. 3r = 6
i.e. r = 2
i.e. coefficient x is 5C2 a3 = 10 a3

Question 17:

If A and B are the coefficient of xn in the expansion of (1 + x)2n and (1 + x)2n – 1 respectively, then $\frac{A}{B}=$___________.

The coefficient of xn is (1 + x)2n is ?
Since Tr +1 = 2nCr xr
For xn coefficient put r = n
i.e coefficient of xn is 2nC
i.e. A = 2nCn
and for coefficient of xn in (1 + x)2n–1
Tr +1 = 2n–1Cr xr
Put r = n
i.e. coefficient of xn in (1 + x)2n–1 is 2n–1Cn
i.e B = 2n–1Cn

Question 18:

If the coefficients of x7 and x8 in ${\left(2+\frac{x}{3}\right)}^{n}$are equal, then n = ___________.

According to given condition,
Coefficient of x7 = Coefficient of x8

Question 19:

If 13th term in the expansion of ${\left({x}^{2}+\frac{2}{x}\right)}^{n}$ is independent of x, then the value of n is ___________.

If for r = 12, i.e 13th term is independent of x
2n – 3r = 0
⇒ 2n = 3 × 12
i.e. n = 18

Question 20:

The term independent of x in the expansion of ${\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^{10}$ is ___________.

For tern to be independent of x, 5 – r = 0
i.e r = 5

Question 1:

Write the number of terms in the expansion of ${\left(2+\sqrt{3}x\right)}^{10}+{\left(2-\sqrt{3}x\right)}^{10}$.

Question 2:

Write the sum of the coefficients in the expansion of ${\left(1-3x+{x}^{2}\right)}^{111}$.

Question 3:

Write the number of terms in the expansion of ${\left(1-3x+3{x}^{2}-{x}^{3}\right)}^{8}$.

Question 4:

Write the middle term in the expansion of $\left(\frac{2{x}^{2}}{3}+\frac{3}{2{x}^{2}}\right)$.

Question 5:

Which term is independent of x, in the expansion of ${\left(x-\frac{1}{3{x}^{2}}\right)}^{9}?$

Question 6:

If a and b denote respectively the coefficients of xm and xn in the expansion of ${\left(1+x\right)}^{m+n}$, then write the relation between a and b.

Question 7:

If a and b are coefficients of xn in the expansions of respectively, then write the relation between a and b.

Question 8:

Write the middle term in the expansion of ${\left(x+\frac{1}{x}\right)}^{10}$.

Question 9:

If a and b denote the sum of the coefficients in the expansions of ${\left(1-3x+10{x}^{2}\right)}^{n}$ and ${\left(1+{x}^{2}\right)}^{n}$ respectively, then write the relation between a and b.

$\mathrm{Here},\phantom{\rule{0ex}{0ex}}a=1-3+10=8={2}^{3}\phantom{\rule{0ex}{0ex}}b=1+1=2\phantom{\rule{0ex}{0ex}}⇒a={b}^{3}$

Question 10:

Write the coefficient of the middle term in the expansion of ${\left(1+x\right)}^{2n}$.

Question 11:

Write the number of terms in the expansion of ${\left[{\left(2x+{y}^{3}\right)}^{4}\right]}^{7}$.

In the binomial expansion of ${\left(a+b\right)}^{n}$, total number of terms will be (n + 1).

Now, ${\left[{\left(2x+{y}^{3}\right)}^{4}\right]}^{7}={\left(2x+{y}^{3}\right)}^{28}$

Therefore, in the expansion of ${\left[{\left(2x+{y}^{3}\right)}^{4}\right]}^{7}$, total number of terms will be 28 + 1 = 29.

Question 12:

Find the sum of the coefficients of two middle terms in the binomial expansion of ${\left(1+x\right)}^{2n-1}$.

Hence, the sum of the coefficients of two middle terms in the binomial expansion of ${\left(1+x\right)}^{2n-1}$ is ${}^{2n}{C}_{n}$.

Question 13:

Find the ratio of the coefficients of xp and xq in the expansion of ${\left(1+x\right)}^{p+q}$.

Coefficient of xp in the expansion of ${\left(1+x\right)}^{p+q}$ is ${}^{p+q}{C}_{p}$.

Coefficient of xq in the expansion of ${\left(1+x\right)}^{p+q}$ is ${}^{p+q}{C}_{q}$.

Now,
$\frac{{}^{p+q}{C}_{p}}{{}^{p+q}{C}_{q}}=\frac{\frac{\left(p+q\right)!}{p!q!}}{\frac{\left(p+q\right)!}{q!p!}}=1$

Hence, the ratio of the coefficients of xp and xq in the expansion of ${\left(1+x\right)}^{p+q}$ is 1 : 1.

Question 14:

Write last two digits of the number 3400.

Hence, last two digits of the number 3400 is 01.

Question 15:

Find the number of terms in the expansion of ${\left(a+b+c\right)}^{n}$.

We have,

Further, expanding each term of R.H.S., we note that
First term consists of 1 term.
Second term on simplification gives 2 terms.
Third term on expansion gives 3 terms.
Similarly, fourth term on expansion gives 4 terms and so on.

∴ The total number of terms = 1 + 2 + 3 + .... + (n + 1) = $\frac{\left(n+1\right)\left(n+2\right)}{2}$.

Question 16:

If a and b are the coefficients of xn in the expansion of  respectively, find $\frac{a}{b}$.

Coefficients of xn in the expansion of ${\left(1+x\right)}^{2n}$ is ${}^{2n}{C}_{n}=a$.

Coefficients of xn in the expansion of ${\left(1+x\right)}^{2n-1}$ is ${}^{2n-1}{C}_{n}=b$.

Now,

Hence, $\frac{a}{b}=2$.

Question 17:

Write the total number of terms in the expansion of ${\left(x+a\right)}^{100}+{\left(x-a\right)}^{100}$.

The total number of terms are 101 of which 50 terms get cancelled.

Hence, the total number of terms in the expansion of ${\left(x+a\right)}^{100}+{\left(x-a\right)}^{100}$ is 51.

Question 18:

If , find the value of ${a}_{0}+{a}_{2}+{a}_{4}+...+{a}_{2n}$.

Putting x = 1 and −1 in

we get,

and

Adding (1) and (2), we get
${3}^{n}+1=2\left({a}_{0}+{a}_{2}+...+{a}_{2n}\right)$

Hence, the value of ${a}_{0}+{a}_{2}+{a}_{4}+...+{a}_{2n}$ is $\frac{{3}^{n}+1}{2}$.

View NCERT Solutions for all chapters of Class 11