Rd Sharma Xi 2018 Solutions for Class 12 Science Math Chapter 21 Some Special Series are provided here with simple step-by-step explanations. These solutions for Some Special Series are extremely popular among Class 12 Science students for Math Some Special Series Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 12 Science Math Chapter 21 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

#### Question 1:

13 + 33 + 53 + 73 + ...

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}={\left(2n-1\right)}^{3}$

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

#### Question 2:

22 + 42 + 62 + 82 + ...

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}={\left(2n\right)}^{2}$

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

#### Question 3:

1.2.5 + 2.3.6 + 3.4.7 + ...

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

#### Question 4:

1.2.4 + 2.3.7 +3.4.10 + ...

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}=n\left(n+1\right)\left(3n+1\right)=n\left(3{n}^{2}+4n+1\right)=\left(3{n}^{3}+4{n}^{2}+n\right)$

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

#### Question 5:

1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:
${T}_{n}=1+2+3+4+5+...+n=\frac{n\left(n+1\right)}{2}=\frac{{n}^{2}+n}{2}$

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

$⇒{S}_{n}=\sum _{k=1}^{n}\left(\frac{{k}^{2}+k}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{1}{2}\sum _{k=1}^{n}\left({k}^{2}+k\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{1}{2}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)}{4}\left(\frac{2n+1}{3}+1\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)}{4}\left(\frac{2n+4}{3}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)\left(2n+4\right)}{12}\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)\left(n+2\right)}{6}$

#### Question 6:

1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}=n\left(n+1\right)={n}^{2}+n$

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

$⇒{S}_{n}=\sum _{k=1}^{n}\left({k}^{2}+k\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\sum _{k=1}^{n}{k}^{2}+\sum _{k=1}^{n}k\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)}{2}\left(\frac{2n+1}{3}+1\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)}{2}\left(\frac{2n+4}{3}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)\left(2n+4\right)}{6}\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)\left(n+2\right)}{3}$

#### Question 7:

3 × 12 + 5 ×22 + 7 × 32 + ...

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}=\left(2n+1\right){n}^{2}=2{n}^{3}+{n}^{2}$

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

$⇒{S}_{n}=\sum _{k=1}^{n}\left(2{k}^{3}+{k}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\underset{k=1}{\overset{n}{2\sum }}{k}^{3}+\sum _{k=1}^{n}{k}^{2}\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\left[\frac{2{n}^{2}{\left(n+1\right)}^{2}}{4}+\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\left[\frac{{n}^{2}{\left(n+1\right)}^{2}}{2}+\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)}{2}\left[n\left(n+1\right)+\frac{2n+1}{3}\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)}{2}\left(\frac{3{n}^{2}+3n+2n+1}{3}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)}{2}\left(\frac{3{n}^{2}+5n+1}{3}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)}{6}\left(3{n}^{2}+5n+1\right)$

#### Question 8:

Find the sum of the series whose nth term is:
(i) 2n2 − 3n + 5
(ii) 2n3 + 3n2 − 1
(iii) n3 − 3n
(iv) n (n + 1) (n + 4)
(v) (2n − 1)2

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

(i)
${T}_{n}=2{n}^{2}-3n+5$

Let ${S}_{n}$ be the sum of n terms of the given series.

Now,

(ii)
${T}_{n}=2{n}^{3}+3{n}^{2}-1$

Let ${S}_{n}$ be the sum of n terms of the given series.

Now,

(iii)
${T}_{n}={n}^{3}-{3}^{n}$

Let ${S}_{n}$ be the sum of n terms of the given series.

Now,

$⇒{S}_{n}=\sum _{k=1}^{n}\left({k}^{3}-{3}^{k}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\sum _{k=1}^{n}{k}^{3}-\sum _{k=1}^{n}{3}^{k}\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{{n}^{2}{\left(n+1\right)}^{2}}{4}-\left(3+{3}^{2}+{3}^{3}+{3}^{4}+...+{3}^{n}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{{n}^{2}{\left(n+1\right)}^{2}}{4}-\left[\frac{3\left({3}^{n}-1\right)}{3-1}\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{{n}^{2}{\left(n+1\right)}^{2}}{4}-\frac{3}{2}\left({3}^{n}-1\right)$

(iv)
${T}_{n}=n\left(n+1\right)\left(n+4\right)=\left({n}^{2}+n\right)\left(n+4\right)={n}^{3}+5{n}^{2}+4n$

Let ${S}_{n}$ be the sum of n terms of the given series.

Now,

(v)
${T}_{n}={\left(2n-1\right)}^{2}$

Let ${S}_{n}$ be the sum of n terms of the given series.

Now,

#### Question 9:

Find the 20th term and the sum of 20 terms of the series 2 × 4 + 4 × 6 + 6 × 8 + ...

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}=2n\left(2n+2\right)=4{n}^{2}+4n$

For n = 20, we have:

Therefore, the 20th term of the given series is 1680.

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

$⇒{S}_{n}=\sum _{k=1}^{n}\left(4{k}^{2}+4k\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\underset{k=1}{\overset{n}{4\sum }}{k}^{2}+4\sum _{k=1}^{n}k$

For n = 20, we have:

Hence, the sum of the first 20 terms of the series is 12320.

#### Question 1:

3 + 5 + 9 + 15 + 23 + ...

Let ${T}_{n}$ be the nth term and ${S}_{n}$ be the sum to n terms of the given series.

Thus, we have:

...(1)

Equation (1) can be rewritten as:

...(2)

On subtracting (2) from (1), we get:

The sequence of difference of successive terms is 2, 4, 6, 8,...
We observe that it is an AP with common difference 2 and first term 2.

Thus, we have:

$3+\left[\frac{\left(n-1\right)}{2}\left\{4+\left(n-2\right)2\right\}\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒3+\left[\frac{\left(n-1\right)}{2}\left(2n\right)\right]={T}_{n}\phantom{\rule{0ex}{0ex}}⇒3+n\left(n-1\right)={T}_{n}$

Now,

#### Question 2:

2 + 5 + 10 + 17 + 26 + ...

Let ${T}_{n}$ be the nth term and ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

...(1)

Equation (1) can be rewritten as:

...(2)

On subtracting (2) from (1), we get:

The sequence of difference of successive terms is 3, 5, 7, 9,...

We observe that it is an AP with common difference 2 and first term 3.

Thus, we have:

$2+\left[\frac{\left(n-1\right)}{2}\left\{6+\left(n-2\right)2\right\}\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒2+\left[{n}^{2}-1\right]={T}_{n}\phantom{\rule{0ex}{0ex}}⇒\left[{n}^{2}+1\right]={T}_{n}$

Now,

#### Question 3:

1 + 3 + 7 + 13 + 21 + ...

Let ${T}_{n}$ be the nth term and ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

...(1)

Equation (1) can be rewritten as:

...(2)

On subtracting (2) from (1), we get:

The sequence of difference of successive terms is 2, 4, 6, 8,...

We observe that it is an AP with common difference 2 and first term 2.

Thus, we have:

$1+\left[\frac{\left(n-1\right)}{2}\left\{4+\left(n-2\right)2\right\}\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒1+\left[{n}^{2}-n\right]={T}_{n}\phantom{\rule{0ex}{0ex}}⇒\left[{n}^{2}-n+1\right]={T}_{n}$

Now,

#### Question 4:

3 + 7 + 14 + 24 + 37 + ...

Let ${T}_{n}$ be the nth term and ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

...(1)

Equation (1) can be rewritten as:

...(2)

On subtracting (2) from (1), we get:

The sequence of difference of successive terms is 4, 7, 10, 13,...

We observe that it is an AP with common difference 3 and first term 4.

Thus, we have:

$3+\left[\frac{\left(n-1\right)}{2}\left\{8+\left(n-2\right)3\right\}\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒3+\left[\frac{\left(n-1\right)}{2}\left(3n+2\right)\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒\left[\frac{3{n}^{2}-n+4}{2}\right]={T}_{n}\phantom{\rule{0ex}{0ex}}⇒\left[\frac{3}{2}{n}^{2}-\frac{n}{2}+2\right]={T}_{n}$

Now,

#### Question 5:

1 + 3 + 6 + 10 + 15 + ...

Let ${T}_{n}$ be the nth term and ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

....(1)

Equation (1) can be rewritten as:

...(2)

On subtracting (2) from (1), we get:

The sequence of difference of successive terms is 2, 3, 4, 5,...

We observe that it is an AP with common difference 1 and first term 2.

Thus, we have:

$1+\left[\frac{\left(n-1\right)}{2}\left(4+\left(n-2\right)1\right)\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒1+\left[\frac{\left(n-1\right)}{2}\left(n+2\right)\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒\left[\frac{{n}^{2}+n}{2}\right]={T}_{n}$

Now,

#### Question 6:

1 + 4 + 13 + 40 + 121 + ...

Let ${T}_{n}$ be the nth term and ${S}_{n}$ be the sum to n terms of the given series.

Thus, we have:

${S}_{n}=1+4+13+40+121+...+{T}_{n-1}+{T}_{n}$       ...(1)

Equation (1) can be rewritten as:

...(2)

On subtracting (2) from (1), we get:

The sequence of difference between successive terms is 3, 9, 27, 81,...

We observe that it is a GP with common ratio 3 and first term 3.

Thus, we have:

$1+\left[\frac{3\left({3}^{n-1}-1\right)}{3-1}\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒1+\left[\frac{\left({3}^{n}-3\right)}{2}\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒\left(\frac{{3}^{n}}{2}-\frac{1}{2}\right)-{T}_{n}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\left(\frac{{3}^{n}}{2}-\frac{1}{2}\right)={T}_{n}$

#### Question 7:

4 + 6 + 9 + 13 + 18 + ...

Let ${T}_{n}$ be the nth term and ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

${S}_{n}=4+6+9+13+18+...+{T}_{n-1}+{T}_{n}$       ...(1)

Equation (1) can be rewritten as:

...(2)

On subtracting (2) from (1), we get:

The sequence of difference between successive terms is 2, 3, 4, 5,...

We observe that it is an AP with common difference 1 and first term 2.

Now,

$4+\left[\frac{\left(n-1\right)}{2}\left\{4+\left(n-2\right)1\right\}\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒4+\left[\frac{\left(n-1\right)}{2}\left(n+2\right)\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒4+\left[\frac{{n}^{2}+n}{2}-1\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒\left[\frac{{n}^{2}}{2}+\frac{n}{2}+3\right]={T}_{n}$

#### Question 8:

2 + 4 + 7 + 11 + 16 + ...

Let ${S}_{n}$ be the sum of n terms and ${T}_{n}$ be the nth term of the given series.

Thus, we have:

...(1)

Equation (1) can be rewritten as:

${S}_{n}=2+4+7+11+16+...+{T}_{n-1}+{T}_{n}$            ...(2)

On subtracting (2) from (1), we get:

$⇒2+\left[\frac{\left(n-1\right)}{2}\left(4+\left(n-2\right)1\right)\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒2+\left[\frac{\left(n-1\right)}{2}\left(n+2\right)\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒2+\left[\frac{{n}^{2}+n}{2}-1\right]-{T}_{n}=0\phantom{\rule{0ex}{0ex}}⇒\left[\frac{{n}^{2}}{2}+\frac{n}{2}+1\right]={T}_{n}$

#### Question 9:

$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

#### Question 10:

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

#### Question 1:

Write the sum of the series 2 + 4 + 6 + 8 + ... + 2n.

${S}_{n}=2+4+6+8+...+2n$

#### Question 2:

Write the sum of the series 12 − 22 + 32 − 42 + 52 − 62 + ... + (2n − 1)2 − (2n)2.

The given series can be rewritten as:

$=-\left[3+7+11+...+\left(4n-1\right)\right]\phantom{\rule{0ex}{0ex}}=-\left[\frac{n}{2}\left\{3×2+\left(n-1\right)4\right\}\right]\phantom{\rule{0ex}{0ex}}=-\left[\frac{n}{2}\left(4n+2\right)\right]\phantom{\rule{0ex}{0ex}}=-n\left(2n+1\right)$

#### Question 3:

Write the sum to n terms of a series whose rth term is r + 2r.

Series whose rth term is r + 2r:

$\left(1+{2}^{1}\right)+\left(2+{2}^{2}\right)+\left(3+{2}^{3}\right)+\left(4+{2}^{4}\right)+...+\left(n+{2}^{n}\right)$

Thus, we have:

If .

#### Question 5:

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then write the value of k.

According to the question,

#### Question 6:

Write the sum of 20 terms of the series $1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....$

Let the nth term be ${a}_{n}$.
Here,
${a}_{n}=\frac{1}{n}\left(1+2+3+...+n\right)=\left(\frac{n+1}{2}\right)$
We know:
${S}_{n}=\sum _{k=1}^{n}{a}_{k}$
Thus, we have:
${S}_{20}=\sum _{k=1}^{20}{a}_{k}$

#### Question 7:

Write the 50th term of the series 2 + 3 + 6 + 11 + 18 + ...

So, the 50th term of the given series is 2403.

#### Question 8:

Let Sn denote the sum of the cubes of first n natural numbers and sn denote the sum of first n natural numbers. Then, write the value of
$\sum _{r=1}^{n}\frac{{S}_{r}}{{s}_{r}}$.

#### Question 1:

The sum to n terms of the series $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+....+....$ is
(a) $\sqrt{2n+1}$

(b) $\frac{1}{2}\sqrt{2n+1}$

(c) $\sqrt{2n+1}-1$

(d) $\frac{1}{2}\left|\sqrt{2n+1}-1\right|$

(d) $\frac{1}{2}\left\{\sqrt{2n+1}-1\right\}$

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:
${T}_{n}=\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}=\frac{\sqrt{2n+1}-\sqrt{2n-1}}{2}$

Now,

Let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

#### Question 2:

The sum of the series is
(a)

(b)

(c)

(d) none of these

(c)

Let

#### Question 3:

The value of is equal to

(a)

(b)

(c)

(d) none of these

(b)

We have:

#### Question 4:

If ∑ n = 210, then ∑ n2 =
(a) 2870
(b) 2160
(c) 2970
(d) none of these

(a) 2870

Given:
n = 210

Now,

#### Question 5:

If Sn = , then Sn is equal to
(a) 2nn − 1

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

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

(d) 2n − 1

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

We have:
Sn =

#### Question 6:

If $1+\frac{1+2}{2}+\frac{1+2+3}{3}+....$ to n terms is S, then S is equal to

(a)

(b)

(c)

(d) n2

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}=\frac{1+2+3+4+5+...+n}{n}=\frac{n\left(n+1\right)}{2n}=\frac{n}{2}+\frac{1}{2}$

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

#### Question 7:

Sum of n terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+$ .... is
(a)

(b) 2n (n + 1)

(c)

(d) 1

(c)

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}=\sqrt{2×{n}^{2}}=n\sqrt{2}$

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

#### Question 8:

The sum of 10 terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+$.... is
(a)

(b)

(c) $\frac{121}{\sqrt{3}-1}$

(d)

(a)

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}=\sqrt{2×{3}^{n-1}}=\sqrt{2}\left(\sqrt{{3}^{n-1}}\right)$

Now, let ${S}_{10}$ be the sum of 10 terms of the given series.

Thus, we have:

#### Question 9:

The sum of the series 12 + 32 + 52 + ... to n terms is
(a)

(b)

(c)

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

(b)

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}={\left(2n-1\right)}^{2}=4{n}^{2}+1-4n$

Now, let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

#### Question 10:

The sum of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+$..... to n terms is
(a) $n-\frac{1}{2}\left({3}^{-n}-1\right)$

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

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

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

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

Let ${T}_{n}$ be the nth term of the given series.

Thus, we have:

${T}_{n}=\frac{{3}^{n}-1}{{3}^{n}}=1-\frac{1}{{3}^{n}}$

Now,

Let ${S}_{n}$ be the sum of n terms of the given series.

Thus, we have:

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