Class 11 RD Sharma Solutions – Chapter 21 Some Special Series- Exercise 21.2

Sum the following series to n terms:

Question 1. 3 + 5 + 9 + 15 + 23 + . . . . n terms

Solution:

We are given the series: 3 + 5 + 9 + 15 + 23 + . . . . n terms.
Let’s take S as the sum of this series. Therefore,
S = 3 + 5 + 9 + 15 + 23 + . . . . + a_n–1+ a_n . . . .(1)
S = 3 + 5 + 9 + 15 + 23 + . . . . + a_n–2 + a_n–1+ a_n . . . .(2)
On subtracting (2) from (1), we get
=> S – S = [3 + (5 + 9 + 15 + 23 + . . . . + a_n–1 + a_n)] – [(3 + 5 + 9 + 15 + 23 + . . . . + a_n–2 + a_n–1) +a_n]
=> 0 = 3 + [(5 – 3) + (9 – 5) + (15 – 9) + . . . . + (a_n – a_n–1)] – a_n
=> a_n = 3 + [2 + 4 + 6 + . . . . (n–1) terms]
As the series 2+ 4 + 6 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 4–2 = 2. So, we get,
=> a_n = 3 + (n–1) [2(2)+(n–2)2]/2
= 3+ (n–1) [4+2n–4]/2
= 3 + n(n–1)
= n²– n+3
Now we have to summate our n^th term to find the sum(S_n) of this series.
$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k + \sum_{k=1}^{n} 3$
= $\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+3n$
= $n\left[\frac{(n+1)(2n+1)-3(n+1)+18}{6}\right]$
= $\frac{n}{6}\left[2n^2+n+2n+1-3n-3+18\right]$
= $\frac{n}{6}\left[2n^2+16\right]$
= $\frac{n(n^2+8)}{3}$
Therefore, sum of the given series up to n terms is $\frac{n(n^2+8)}{3}$ .

Question 2. 2 + 5 + 10 + 17 + 26 + . . . . n terms

Solution:

We are given the series: 2 + 5 + 10 + 17 + 26 + . . . . n terms.
Let’s take S as the sum of this series. Therefore,
S = 2 + 5 + 10 + 17 + 26 + . . . . + a_n–1 + a_n . . . .(1)
S = 2 + 5 + 10 + 17 + 26 + . . . . + a_n–2 + a_n–1 + a_n . . . .(2)
On subtracting (2) from (1), we get
=> S – S = [2 + (5 + 10 + 17 + 26 + . . . . + a_n–1 + a_n)] – [(2 + 5 + 10 + 17 + 26 + . . . . + a_n–2 + a_n–1) +a_n]
=> 0 = 2 + [(5 – 2) + (10 – 5) + (17 – 10) + . . . . + (a_n – a_n–1)] – a_n
=> a_n = 2 + [3 + 5 + 7 + . . . . (n–1) terms]
As the series 3 + 5 + 7 + . . . . (n–1) terms is an A.P., with first term(a) = 3 and common difference(d) = 5–3 = 2. So, we get,
=> a_n = 2 + (n–1) [2(3)+(n–2)2]/2
= 2+ (n–1) [6+2n–4]/2
= 2 + (n–1)(n+1)
= n² – 1 + 2
= n² + 1
Now we have to summate our n^th term to find the sum(S_n) of this series.
$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} 1$
= $\frac{n(n+1)(2n+1)}{6}+n$
= $\frac{n(n+1)(2n+1)+6n}{6}$
= $\frac{n(2n^2+3n+7)}{6}$
Therefore, sum of the given series up to n terms is $\frac{n(2n^2+3n+7)}{6}$ .

Question 3. 1 + 3 + 7 + 13 + 21 + 31 + . . . . n terms

Solution:

We are given the series: 1 + 3 + 7 + 13 + 21 + 31 + . . . . n terms.
Let’s take S as the sum of this series. Therefore,
S = 1 + 3 + 7 + 13 + 21 + 31 + . . . . + a_n–1 + a_n . . . .(1)
S = 1 + 3 + 7 + 13 + 21 + 31 + . . . . + a_n–2 + a_n–1 + a_n . . . .(2)
On subtracting (2) from (1), we get
=> S – S = [1 + (3 + 7 + 13 + 21 + 31 + . . . . + a_n–1 + an)] – [(1 + 3 + 7 + 13 + 21 + 31 + . . . . + a_n–2 + a_n–1) + a_n]
=> 0 = 1 + [(3 – 1) + (7 – 3) + (13 – 7) + . . . . + (a_n – a_n–1)] – a_n
=> a_n = 1 + [2 + 4 + 6 + . . . . (n–1) terms]
As the series 2 + 4 + 6 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 4–2 = 2. So, we get,
=> a_n = 1 + (n–1) [2(2)+(n–2)2]/2
= 1+ (n–1) [4+2n–4]/2
= 1 + n(n–1)
= n² – n+1
Now we have to summate our n^th term to find the sum(S_n) of this series.
$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1$
= $\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+n$
= $\frac{n(n+1)(2n+1)-3n(n+1)+6n}{6}$
= $\frac{n(2n^2+n+2n+1-3n-3+6)}{6}$
= $\frac{n(2n^2+4)}{6}$
= $\frac{n(n^2+2)}{3}$
Therefore, sum of the given series up to n terms is $\frac{n(n^2+2)}{3}$ .

Question 4. 3 + 7 + 14 + 24 + 37 + . . . . n terms

Solution:

We are given the series: 3 + 7 + 14 + 24 + 37 + . . . . n terms.
Let’s take S as the sum of this series. Therefore,
S = 3 + 7 + 14 + 24 + 37 + . . . . + a_n–1 + a_n . . . .(1)
S = 3 + 7 + 14 + 24 + 37 + . . . . + a_n–2 + a_n–1 + a_n . . . .(2)
On subtracting (2) from (1), we get
=> S – S = [3 + (7 + 14 + 24 + 37 + . . . . + a_n–1 + a_n)] – [(3 + 7 + 14 + 24 + 37 + . . . . + a_n–2 + a_n–1) + a_n]
=> 0 = 3 + [(7 – 3) + (14 – 7) + (24 – 14) + . . . . + (a_n – a_n–1)] – a_n
=> a_n = 3 + [4 + 7 + 10 + . . . . (n–1) terms]
As the series 4 + 7 + 10 + . . . . (n–1) terms is an A.P., with first term(a) = 4 and common difference(d) = 7–4 = 3. So, we get,
=> a_n = 3 + (n–1) [2(4)+(n–2)3]/2
= 3+ (n–1) [8+3n–6]/2
= 3 + (n–1)(3n+2)/2
= $\frac{6+(n-1)(3n+2)}{2}$
= $\frac{6+3n^2-n-2}{2}$
= $\frac{3n^2-n+4}{2}$
Now we have to summate our n^th term to find the sum(S_n) of this series.
$S_n = \sum_{k=1}^{n} a_k = \frac{1}{2}\left[\sum_{k=1}^{n} 3k^2 - \sum_{k=1}^{n} k + \sum_{k=1}^{n} 4\right]$
= $\frac{3}{2}\sum_{k=1}^{n} k^2 - \frac{1}{2}\sum_{k=1}^{n} k + \sum_{k=1}^{n} 2$
= $\frac{3}{2}[\frac{n(n+1)(2n+1)}{6}]-\frac{1}{2}[\frac{n(n+1)}{2}]+2n$
= $\frac{n(n+1)(2n+1)-n(n+1)+8n}{4}$
= $\frac{n(2n^2+2n+8)}{4}$
= $\frac{n(n^2+n+4)}{2}$
Therefore, sum of the given series up to n terms is $\frac{n(n^2+n+4)}{2}$ .

Question 5. 1 + 3 + 6 + 10 + 15 + . . . . n terms

Solution:

We are given the series: 1 + 3 + 6 + 10 + 15 + . . . . n terms.
Let’s take S as the sum of this series. Therefore,
S = 1 + 3 + 6 + 10 + 15 + . . . . + a_n–1 + a_n . . . .(1)
S = 1 + 3 + 6 + 10 + 15 + . . . . + a_n–2 + a_n–1 + a_n . . . .(2)
On subtracting (2) from (1), we get
=> S – S = [1 + (3 + 6 + 10 + 15 + . . . . + a_n–1 + a_n)] – [(1 + 3 + 6 + 10 + 15 + . . . . + a_n–2 + a_n–1) + a_n]
=> 0 = 1 + [(3 – 1) + (6 – 3) + (10 – 6) + . . . . + (a_n – a_n–1)] – a_n
=> a_n = 1 + [2 + 3 + 4 + . . . . (n–1) terms]
As the series 2 + 3 + 4 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 3–2 = 1. So, we get,
=> a_n = 1 + (n–1) [2(2)+(n–2)1]/2
= 1+ (n–1) [4+n–2]/2
= 1 + (n–1)(n+2)/2
= $\frac{2+(n-1)(n+2)}{2}$
= $\frac{2+n^2+2n-n-2}{2}$
= $\frac{n^2+n}{2}$
Now we have to summate our n^th term to find the sum(S_n) of this series.
$S_n = \sum_{k=1}^{n} a_k = \frac{1}{2}\left[\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k\right]$
= $\frac{1}{2}\sum_{k=1}^{n} k^2 + \frac{1}{2}\sum_{k=1}^{n} k$
= $\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}\right] + \frac{1}{2}\left[\frac{n(n+1)}{2}\right]$
= $\frac{n(n+1)}{2}\left[\frac{2n+1}{6}+\frac{1}{2}\right]$
= $\frac{n(n+1)(2n+4)}{12}$
= $\frac{n(n+1)(n+2)}{6}$
Therefore, sum of the given series up to n terms is $\frac{n(n+1)(n+2)}{6}$ .

Question 6. 1 + 4 + 13 + 40 + 121 + . . . . n terms

Solution:

We are given the series: 1 + 4 + 13 + 40 + 121 + . . . . n terms.
Let’s take S as the sum of this series. Therefore,
S = 1 + 4 + 13 + 40 + 121 + . . . . + a_n–1 + a_n . . . .(1)
S = 1 + 4 + 13 + 40 + 121 + . . . . + a_n–2 + a_n–1 + a_n . . . .(2)
On subtracting (2) from (1), we get
=> S – S = [1 + (4 + 13 + 40 + 121 + . . . . + a_n–1 + a_n)] – [(1 + 4 + 13 + 40 + 121 + . . . . + a_n–2 + a_n–1) + a_n]
=> 0 = 1 + [(4 – 1) + (13 – 4) + (40 – 13) + . . . . + (a_n – a_n–1)] – a_n
=> a_n = 1 + [3 + 9 + 27 + . . . . (n–1) terms]
As the series 3 + 9 + 27 + . . . . (n–1) terms is a G.P., with first term(a) = 3 and common ratio(r) = 9/3 = 3. So, we get,
=> a_n = 1 + 3(3^n-1–1)/(3–1)
= 1+ 3(3^n-1–1)/2
= 1 + 3ⁿ/2 – 3/2
= 3ⁿ/2 – 1/2
Now we have to summate our n^th term to find the sum(S_n) of this series.
$S_n = \sum_{k=1}^{n} a_k = \frac{1}{2}\sum_{k=1}^{n} 3^n - \frac{1}{2}\sum_{k=1}^{n} 1$
= $\frac{3^1+3^2+3^3+...+3^n}{2}-\frac{n}{2}$
= $\frac{3(3^n-1)}{(3-1)2}-\frac{n}{2}$
= $\frac{3(3^n)-2n-3}{4}$
= $\frac{3^{n+1}-2n-3}{4}$
Therefore, sum of the given series up to n terms is $\frac{3^{n+1}-2n-3}{4}$ .

Question 7. 4 + 6 + 9 + 13 + 18 + . . . . n terms

Solution:

We are given the series: 4 + 6 + 9 + 13 + 18 + . . . . n terms.
Let’s take S as the sum of this series. Therefore,
S = 4 + 6 + 9 + 13 + 18 + . . . . + a_n–1 + a_n . . . .(1)
S = 4 + 6 + 9 + 13 + 18 + . . . . + a_n–2 + a_n–1 + a_n . . . .(2)
On subtracting (2) from (1), we get
=> S – S = [4 + (6 + 9 + 13 + 18 + . . . . + a_n–1 + a_n)] – [(4 + 6 + 9 + 13 + 18 + . . . . + a_n–2 + a_n–1) + a_n]
=> 0 = 4 + [(6 – 4) + (9 – 6) + (13 – 9) + . . . . + (a_n – a_n–1)] – a_n
=> a_n = 4 + [2 + 3 + 4 + . . . . (n–1) terms]
As the series 2 + 3 + 4 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 3–2 = 1. So, we get,
=> a_n= 4 + (n–1) [2(2)+(n–2)1]/2
= 4 + (n–1)(n+2)/2
= $\frac{8+(n-1)(n+2)}{2}$
= $\frac{n^2+n+6}{2}$
Now we have to summate our n^th term to find the sum(S_n) of this series.
$S_n = \sum_{k=1}^{n} a_k = \frac{1}{2}\left[\sum_{k=1}^{n} n^2 + \sum_{k=1}^{n} n + \sum_{k=1}^{n} 6\right]$
= $\frac{1}{2}\sum_{k=1}^{n} n^2 + \frac{1}{2}\sum_{k=1}^{n} n + \sum_{k=1}^{n} 3$
= $\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}] + \frac{1}{2}[\frac{n(n+1)}{2}] + 3n$
= $\frac{n[(n+1)(2n+1)+3(n+1)+36]}{12}$
= $\frac{n[2n^2+6n+40]}{12}$
= $\frac{n[n^2+3n+20]}{6}$
Therefore, sum of the given series up to n terms is $\frac{n[n^2+3n+20]}{6}$ .

Question 8. 2 + 4 + 7 + 11 + 16 + . . . . n terms

Solution:

We are given the series: 2 + 4 + 7 + 11 + 16 + . . . . n terms.
Let’s take S as the sum of this series. Therefore,
S = 2 + 4 + 7 + 11 + 16 + . . . . + a_n–1 + a_n . . . .(1)
S = 2 + 4 + 7 + 11 + 16 + . . . . + a_n–2 + a_n–1 + a_n . . . .(2)
On subtracting (2) from (1), we get
=> S – S = [2 + (4 + 7 + 11 + 16 + . . . . + a_n–1 + a_n)] – [(2 + 4 + 7 + 11 + 16 + . . . . + a_n–2 + a_n–1) + a_n]
=> 0 = 2 + [(4 – 2) + (7 – 4) + (11 – 7) + . . . . + (a_n – a_n–1)] – a_n
=> a_n = 2 + [2 + 3 + 4 + . . . . (n–1) terms]
As the series 2 + 3 + 4 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 3–2 = 1. So, we get,
=> a_n = 2 + (n–1) [2(2)+(n–2)1]/2
= 2 + (n–1)(n+2)/2
= $\frac{4+n^2+2n-n-2}{2}$
= $\frac{n^2+n+2}{2}$
Now we have to summate our n^th term to find the sum(S_n) of this series.
$S_n = \sum_{k=1}^{n} a_k = \frac{1}{2}\left[\sum_{k=1}^{n} n^2 + \sum_{k=1}^{n} n + \sum_{k=1}^{n} 2\right]$
= $\frac{1}{2}\sum_{k=1}^{n} n^2 + \frac{1}{2}\sum_{k=1}^{n} n + \sum_{k=1}^{n} 1$
= $\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}]+\frac{1}{2}[\frac{n(n+1)}{2}]+n$
= $\frac{n[(n+1)(2n+1)+3(n+1)+12n]}{12}$
= $\frac{n(2n^2+6n+16)}{12}$
= $\frac{n(n^2+3n+8)}{6}$
Therefore, sum of the given series up to n terms is $\frac{n(n^2+3n+8)}{6}$ .

Question 9. $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....n\hspace{0.1cm}terms$

Solution:

The n^th term of the given series would be,
a_n = $\frac{1}{(3n-2)(3n+1)}$
= $\frac{1}{3}\left[\frac{1}{3n-2}-\frac{1}{3n+1}\right]$
Now we have to summate our n^th term to find the sum(S_n) of this series.
$S_n = \sum_{k=1}^{n} a_k = \frac{1}{3}\sum_{k=1}^{n} \left[\frac{1}{3n-2}-\frac{1}{3n+1}\right]$
= $\frac{1}{3}\left[(1-\frac{1}{4})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})...(\frac{1}{3n-2}-\frac{1}{3n+1})\right]$
= $\frac{1}{3}\left[1-\frac{1}{3n+1}\right]$
= $\frac{3n}{3(3n+1)}$
= $\frac{n}{3n+1}$
Therefore, sum of the given series up to n terms is $\frac{n}{3n+1}$ .

Question 10. $\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.14}+\frac{1}{14.19}....\frac{1}{(5n-4)(5n+1)}$ to n terms.

Solution:

We are given the n^th term of the given series,
a_n = $\frac{1}{(5n-4)(5n+1)}$
= $\frac{1}{5}\left[\frac{1}{5n-4}-\frac{1}{5n+1}\right]$
Now we have to summate our n^th term to find the sum(S_n) of this series.
$S_n = \sum_{k=1}^{n} a_k = \frac{1}{5}\sum_{k=1}^{n} \left[\frac{1}{5n-4}-\frac{1}{5n+1}\right]$
= $\frac{1}{3}\left[(1-\frac{1}{6})+(\frac{1}{6}-\frac{1}{11})+(\frac{1}{11}-\frac{1}{14})...(\frac{1}{5n-4}-\frac{1}{5n+1})\right]$
= $\frac{1}{5}\left[1-\frac{1}{5n+1}\right]$
= $\frac{5n}{5(5n+1)}$
= $\frac{n}{5n+1}$
Therefore, sum of the given series up to n terms is $\frac{n}{5n+1}$ .