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#### 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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