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

#### Answer:

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

#### Answer:

(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 ${\left(\sqrt{3}+\sqrt{2}\right)}^{4}-{\left(\sqrt{3}-\sqrt{2}\right)}^{4}$.

#### Answer:

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

#### Answer:

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

#### Answer:

(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 .

#### Answer:

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

#### Answer:

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.

#### Answer:

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.

#### Answer:

We have,

Thus, â€‹ ${2}^{4n+4}-15n-16$, 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 ${\left(2x-\frac{1}{{x}^{2}}\right)}^{25}$.

#### Answer:

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 ${\left(3{x}^{2}-\frac{1}{{x}^{3}}\right)}^{10}$.

#### Answer:

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 ${\left(3x-\frac{1}{{x}^{2}}\right)}^{10}$

#### Answer:

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 .

#### Answer:

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 ${\left(\frac{4x}{5}+\frac{5}{2x}\right)}^{8}$.

#### Answer:

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 ${\left(x+\frac{2}{x}\right)}^{9}$.

#### Answer:

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 ${\left(\frac{4x}{5}-\frac{5}{2x}\right)}^{8}$.

#### Answer:

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 ${\left(2{x}^{2}-\frac{3}{2x}\right)}^{8}$

#### Answer:

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 .

#### Answer:

(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?

#### Answer:

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?

#### Answer:

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.

#### Answer:

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

#### Answer:

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

#### Answer:

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

#### Answer:

(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.

#### Answer:

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 $p\ne q$.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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 $\frac{1}{6}$.

#### Answer:

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.

#### Answer:

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:

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.

#### Answer:

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

#### 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 ${\left[{\left(2x+{y}^{3}\right)}^{4}\right]}^{7}$.

#### Answer:

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 ${\left(1+x\right)}^{2n-1}$.

#### Answer:

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 ${\left(1+x\right)}^{p+q}$.

#### Answer:

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.

#### Answer:

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

#### Question 15:

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

#### Answer:

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 $\frac{a}{b}$.

#### Answer:

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 ${\left(x+a\right)}^{100}+{\left(x-a\right)}^{100}$.

#### Answer:

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 ${a}_{0}+{a}_{2}+{a}_{4}+...+{a}_{2n}$.

#### Answer:

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 $\frac{{3}^{n}+1}{2}$.

#### 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

#### Answer:

(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

#### Answer:

(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

#### Answer:

(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

#### Answer:

(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

#### Answer:

(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

#### Answer:

(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

#### Answer:

(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

#### Answer:

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

#### Answer:

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

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