Find the sum of the following series to $n$ terms:

Question 1: $3+5+9+15+23 + \ldots$

Let $T_n$ be the $n^{th}$ term and $S_n$  be the sum of the $n$ terms.

$S_n = 3+5+9+15+23 + \ldots + T_{n-1}+ T_n$     … … … … … i)

$S_n = \hspace{0.75cm} 3+5+9+15+23 + \ldots + T_{n-1}+ T_n$     … … … … … ii)

Subtracting ii) from i) we get

$0 = 3 + [ 2+4+6+ \ldots + (T_n - T_{n-1}) ] - T_n$

$\Rightarrow 0 = 3 +$ $\frac{(n-1)}{2}$ $\Big[ 2(2) + ( n - 1 - 1) (2) \Big] - T_n$

$\Rightarrow T_n = 3 +$ $\frac{(n-1)}{2}$ $\Big[ 4 + ( n - 2)( 2) \Big]$

$\Rightarrow T_n = 3 +$ $\frac{(n-1)}{2}$ $(2n)$

$\Rightarrow T_n = 3 + n ( n-1)$

$\Rightarrow T_n = n^2 - n + 3$

$\therefore S_n = \sum \limits_{k=1}^n (k^2 -k+3)$

$\Rightarrow S_n = \sum \limits_{k=1}^n k^2 - \sum \limits_{k=1}^n k + 3 \sum \limits_{k=1}^n 1$

$\Rightarrow S_n =$ $\frac{n(n+1)(2n+1)}{6}$ $-$ $\frac{n(n+1)}{2}$ $+ 3n$

$\Rightarrow S_n =$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{2n+1}{3}$ $- 1 \Big] + 3n$

$\Rightarrow S_n =$ $\frac{n(n+1)}{2}$ $\frac{(2n-2)}{3}$ $+ 3n$

$\Rightarrow S_n =$ $\frac{n(n^2-1)}{3}$ $+ 3n$

$\Rightarrow S_n =$ $\frac{n^3- n + 9n}{3}$

$\Rightarrow S_n =$ $\frac{n^3 + 8n}{3}$

$\Rightarrow S_n =$ $\frac{n}{3}$ $(n^2 + 8)$

$\\$

Question 2: $2+5+10+17+26+ \ldots$

Let $T_n$ be the $n^{th}$ term and $S_n$  be the sum of the $n$ terms.

$S_n = 2+5+10+17+26 + \ldots + T_{n-1}+ T_n$     … … … … … i)

$S_n = \hspace{0.75cm} 2+5+10+17+26 + \ldots + T_{n-1}+ T_n$     … … … … … ii)

Subtracting ii) from i) we get

$0 = 2 + [ 3+5+7+9+ \ldots + (T_n - T_{n-1}) ] - T_n$

$\Rightarrow 0 = 2 +$ $\frac{(n-1)}{2}$ $\Big[ 2(3) + ( n - 1 - 1) (2) \Big] - T_n$

$\Rightarrow T_n = 2 +$ $\frac{(n-1)}{2}$ $( 2n+2 )$

$\Rightarrow T_n =$ $\frac{2n^2-2n+2n-2+4}{2}$

$\Rightarrow T_n =$ $\frac{2n^2+2}{2}$

$\Rightarrow T_n = n^2+1$

$\therefore S_n = \sum \limits_{k=1}^n (k^2 +1)$

$\Rightarrow S_n = \sum \limits_{k=1}^n k^2 + \sum \limits_{k=1}^n 1$

$\Rightarrow S_n =$ $\frac{n(n+1)(2n+1)}{6}$ $+ n$

$\Rightarrow S_n =$ $\frac{n(n+1)(2n+1)+6n}{6}$

$\Rightarrow S_n =$ $\frac{n(2n^2+3n+7)}{6}$

$\\$

Question 3: $1+3+7+13+21+ \ldots$

Let $T_n$ be the $n^{th}$ term and $S_n$  be the sum of the $n$ terms.

$S_n = 1+3+7+13+21 + \ldots + T_{n-1}+ T_n$     … … … … … i)

$S_n = \hspace{0.75cm} 1+3+7+13+21 + \ldots + T_{n-1}+ T_n$     … … … … … ii)

Subtracting ii) from i) we get

$0 = 1 + [ 2+4+6+8+ \ldots + (T_n - T_{n-1}) ] - T_n$

$\Rightarrow 0 = 1 +$ $\frac{(n-1)}{2}$ $\Big[ 2(2) + ( n - 1 - 1) (2) \Big] - T_n$

$\Rightarrow T_n = 1 +$ $\frac{(n-1)}{2}$ $(2n)$

$\Rightarrow T_n = 1 + +n(n-1)$

$\Rightarrow T_n = n^2 - n + 1$

$\therefore S_n = \sum \limits_{k=1}^n (k^2 -k+1)$

$\Rightarrow S_n = \sum \limits_{k=1}^n k^2 - \sum \limits_{k=1}^n k + \sum \limits_{k=1}^n 1$

$\Rightarrow S_n =$ $\frac{n(n+1)(2n+1)}{6}$ $-$ $\frac{n(n+1)}{2}$ $+ n$

$\Rightarrow S_n =$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{2n+1}{3}$ $- 1 \Big] + n$

$\Rightarrow S_n =$ $\frac{n(n+1)}{2}$ $\frac{(2n-2)}{3}$ $+ n$

$\Rightarrow S_n =$ $\frac{n(n^2-1)}{3}$ $+ n$

$\Rightarrow S_n =$ $\frac{n^3- n + 3n}{3}$

$\Rightarrow S_n =$ $\frac{n^3 + 2n}{3}$

$\Rightarrow S_n =$ $\frac{n}{3}$ $(n^2 + 2)$

$\\$

Question 4: $3+7+14+24+37+ \ldots$

Let $T_n$ be the $n^{th}$ term and $S_n$  be the sum of the $n$ terms.

$S_n = 3+7+14+24+37 + \ldots + T_{n-1}+ T_n$     … … … … … i)

$S_n = \hspace{0.75cm} 3+7+14+24+37 + \ldots + T_{n-1}+ T_n$     … … … … … ii)

Subtracting ii) from i) we get

$0 = 3 + [ 4+7+10+13 + \ldots + (T_n - T_{n-1}) ] - T_n$

$\Rightarrow 0 = 3 +$ $\frac{(n-1)}{2}$ $\Big[ 2(4) + ( n - 1 - 1) (3) \Big] - T_n$

$\Rightarrow T_n = 3 +$ $\frac{(n-1)}{2}$ $\Big[ 8 + 3n-6 \Big]$

$\Rightarrow T_n = 3 +$ $\frac{(n-1)}{2}$ $(3n+2)$

$\Rightarrow T_n =$ $\frac{3n^2-3n+2n-2+6}{2}$

$\Rightarrow T_n =$ $\frac{3n^2-n+4}{2}$

$\therefore S_n = \sum \limits_{k=1}^n \Big($ $\frac{3k^2-k+4}{2}$ $\Big)$

$\Rightarrow S_n = \frac{3}{2} \sum \limits_{k=1}^n k^2 - \frac{1}{2}\sum \limits_{k=1}^n k + 2 \sum \limits_{k=1}^n 1$

$\Rightarrow S_n =$ $\frac{3}{2} \frac{n(n+1)(2n+1)}{6}$ $-$ $\frac{1}{2} \frac{n(n+1)}{2}$ $+ 2n$

$\Rightarrow S_n =$ $\frac{n(n+1)(2n+1)}{4}$ $-$ $\frac{n(n+1)}{4}$ $+ 2n$

$\Rightarrow S_n =$ $\frac{n(n+1)}{4}$ $[(2n+1)-1 ] + 2n$

$\Rightarrow S_n =$ $\frac{n^2(n+1)}{2}$ $+ 2n$

$\Rightarrow S_n =$ $\frac{n}{2}$ $[ n(n+1) + 4]$

$\Rightarrow S_n =$ $\frac{n}{2}$ $(n^2+n+4)$

$\\$

Question 5: $1+3+6+10+15 + \ldots$

Let $T_n$ be the $n^{th}$ term and $S_n$  be the sum of the $n$ terms.

$S_n = 1+3+6+10+15 + \ldots + T_{n-1}+ T_n$     … … … … … i)

$S_n = \hspace{0.75cm} 1+3+6+10+15 + \ldots + T_{n-1}+ T_n$     … … … … … ii)

Subtracting ii) from i) we get

$0 = 1 + [ 2+3+4+5+ \ldots + (T_n - T_{n-1}) ] - T_n$

$\Rightarrow 0 = 1 +$ $\frac{(n-1)}{2}$ $\Big[ 2(2) + ( n - 1 - 1) (1) \Big] - T_n$

$\Rightarrow T_n = 1 +$ $\frac{(n-1)}{2}$ $( n+2 )$

$\Rightarrow T_n =$ $\frac{2+n^2-n+2n-2}{2}$

$\Rightarrow T_n =$ $\frac{n^2+n}{2}$

$\Rightarrow T_n =$ $\frac{n^2}{2}$ $+$ $\frac{n}{2}$

$\therefore S_n = \sum \limits_{k=1}^n ($ $\frac{k^2}{2}$ $+$ $\frac{k}{2}$ $)$

$\Rightarrow S_n =$ $\frac{1}{2}$ $\sum \limits_{k=1}^n k^2 +$ $\frac{1}{2}$ $\sum \limits_{k=1}^n k$

$\Rightarrow S_n =$ $\frac{1}{2}$ $\frac{n(n+1)(2n+1)}{6}$ $+$ $\frac{1}{2}$ $\frac{n(n+1)}{2}$

$\Rightarrow S_n =$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{2n+1}{6}$ $+$ $\frac{1}{2}$ $\Big]$

$\Rightarrow S_n =$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{2n+4}{6}$ $\Big]$

$\Rightarrow S_n =$ $\frac{n(n+1)(2n+2)}{6}$

$\\$

Question 6: $1+4+13+40+121 + \ldots$

Let $T_n$ be the $n^{th}$ term and $S_n$  be the sum of the $n$ terms.

$S_n = 1+4+13+40+121 + \ldots + T_{n-1}+ T_n$     … … … … … i)

$S_n = \hspace{0.75cm} 1+4+13+40+121 + \ldots + T_{n-1}+ T_n$     … … … … … ii)

Subtracting ii) from i) we get

$0 = 1 + [ 3+9+27+81 + \ldots + (T_n - T_{n-1}) ] - T_n$

$T_n = 1 +$ $\frac{3(3^n-1)}{3-1}$ $=$ $\frac{3^n - 3 +2}{2}$ $=$ $\frac{3^n}{2}$ $-$ $\frac{1}{2}$

$\therefore S_n = \sum \limits_{k=1}^n$ $\frac{3^k}{2}$ $-$ $\frac{1}{2}$

$\Rightarrow S_n =$ $\frac{1}{2}$ $\sum \limits_{k=1}^n 3^k -$ $\frac{1}{2}$ $\sum \limits_{k=1}^n 1$

$\Rightarrow S_n =$ $\frac{1}{2}$ $( 3 + 3^2 + 3^3 + \ldots + 3^n) -$ $\frac{n}{2}$

$\Rightarrow S_n =$ $\frac{1}{2}$ $($ $\frac{3(3^n-1)}{3-1}$ $) -$ $\frac{n}{2}$

$\Rightarrow S_n =$ $\frac{3^{n+1}-3}{4}$ $-$ $\frac{n}{2}$

$\Rightarrow S_n =$ $\frac{3^{n+1} - 2n-3}{4}$

$\\$

Question 7: $4+6+9+13+18+ \ldots$

Let $T_n$ be the $n^{th}$ term and $S_n$  be the sum of the $n$ terms.

$S_n = 4+6+9+13+18 + \ldots + T_{n-1}+ T_n$     … … … … … i)

$S_n = \hspace{0.75cm} 4+6+9+13+18 + \ldots + T_{n-1}+ T_n$     … … … … … ii)

Subtracting ii) from i) we get

$0 = 4 + [ 2+3+4+5 + \ldots + (T_n - T_{n-1}) ] - T_n$

$\Rightarrow 0 = 4 +$ $\frac{(n-1)}{2}$ $\Big[ 2(2) + ( n - 1 - 1) (1) \Big] - T_n$

$\Rightarrow T_n = 4 +$ $\frac{(n-1)}{2}$ $\Big[ n+2 \Big]$

$\Rightarrow T_n =$ $\frac{n^2-n+2n-2+8}{2}$

$\Rightarrow T_n =$ $\frac{n^2+n+6}{2}$

$\therefore S_n = \sum \limits_{k=1}^n \Big($ $\frac{k^2+k+6}{2}$ $\Big)$

$\Rightarrow S_n =$ $\frac{1}{2}$ $\sum \limits_{k=1}^n k^2 +$ $\frac{1}{2}$ $\sum \limits_{k=1}^n k + 3 \sum \limits_{k=1}^n 1$

$\Rightarrow S_n =$ $\frac{1}{2} \frac{n(n+1)(2n+1)}{6}$ $+$ $\frac{1}{2} \frac{n(n+1)}{2}$ $+ 3n$

$\Rightarrow S_n =$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{2n+1}{6}$ $+$ $\frac{1}{2}$ $\Big] + 3n$

$\Rightarrow S_n =$ $\frac{n(n+1)}{2}$ $\Big[ \frac{2n+4}{6} \Big] + 3n$

$\Rightarrow S_n =$ $\frac{n(n+1)(n+2)}{6}$ $+ 3n$

$\Rightarrow S_n =$ $\frac{n}{6}$ $[ n(n+1)(n+2) + 18]$

$\Rightarrow S_n =$ $\frac{n}{6}$ $(n^2+3n+20)$

$\\$

Question 8: $2+4+7+11+16 + \ldots$

Let $T_n$ be the $n^{th}$ term and $S_n$  be the sum of the $n$ terms.

$S_n = 2+4+7+11+16 + \ldots + T_{n-1}+ T_n$     … … … … … i)

$S_n = \hspace{0.75cm} 2+4+7+11+16 + \ldots + T_{n-1}+ T_n$     … … … … … ii)

Subtracting ii) from i) we get

$0 = 2 + [ 2+3+4+5 + \ldots + (T_n - T_{n-1}) ] - T_n$

$\Rightarrow 0 = 2 +$ $\frac{(n-1)}{2}$ $\Big[ 2(2) + ( n - 1 - 1) (1) \Big] - T_n$

$\Rightarrow T_n = 2 +$ $\frac{(n-1)}{2}$ $\Big[ n+2 \Big]$

$\Rightarrow T_n =$ $\frac{n^2-n+2n-2+4}{2}$

$\Rightarrow T_n =$ $\frac{n^2+n+2}{2}$

$\therefore S_n = \sum \limits_{k=1}^n \Big($ $\frac{k^2+k+2}{2}$ $\Big)$

$\Rightarrow S_n =$ $\frac{1}{2}$ $\sum \limits_{k=1}^n k^2 +$ $\frac{1}{2}$ $\sum \limits_{k=1}^n k + \sum \limits_{k=1}^n 1$

$\Rightarrow S_n =$ $\frac{1}{2} \frac{n(n+1)(2n+1)}{6}$ $+$ $\frac{1}{2} \frac{n(n+1)}{2}$ $+ n$

$\Rightarrow S_n =$ $\frac{n(n+1)}{4}$ $\Big[$ $\frac{2n+1}{3}$ $+ 1 \Big] + n$

$\Rightarrow S_n =$ $\frac{n(n+1)}{4}$ $\Big[$ $\frac{2n+4}{3}$ $\Big] + n$

$\Rightarrow S_n =$ $\frac{n(n+1)(n+2)+6n}{6}$

$\Rightarrow S_n =$ $\frac{n}{6}$ $(n^2+3n+8)$

$\\$

Question 9: $\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \ldots$

Given series: $\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + \ldots$

$=$ $\frac{1}{(3 \times 1 - 2)(3 \times 1 + 1)} + \frac{1}{(3 \times 2 - 2)(3 \times 2 + 1)} + \frac{1}{(3 \times 3 - 2)(3 \times 3 + 1)}$ $+ \ldots$

$T_n =$ $\frac{1}{(3n-2)(3n+1)}$

$S_n = \sum \limits_{k=1}^n$ $\frac{1}{(3k-2)(3k+1)}$

$=$ $\frac{1}{3}$ $\Big[ \sum \limits_{k=1}^n \Big($ $\frac{1}{3k-2}$ $-$ $\frac{1}{3k+1}$ $\Big) \Big]$

$=$ $\frac{1}{3}$ $\Big[ \sum \limits_{k=1}^n$ $\frac{1}{3k-2}$ $- \sum \limits_{k=1}^n$ $\frac{1}{3k+1}$ $\Big]$

$=$ $\frac{1}{3}$ $\Big[ \Big( 1 +$ $\frac{1}{4} + \frac{1}{7} + \ldots + \frac{1}{3n-2}$ $\Big) - \Big($ $\frac{1}{4} + \frac{1}{7} + \ldots + \frac{1}{3n-2} + \frac{1}{3n+1}$ $\Big) \Big]$

$=$ $\frac{1}{3}$ $\Big[ 1 -$ $\frac{1}{3n+1}$ $\Big]$

$=$ $\frac{n}{3n+1}$

$\\$

Question 10:  $\frac{1}{1.6} + \frac{1}{6.11} + \frac{1}{11.16} + \frac{1}{16.21} + \ldots + \frac{1}{(5n-4)(5n+1)}$

Given series: $\frac{1}{1.6} + \frac{1}{6.11} + \frac{1}{11.16} + \frac{1}{16.21} + \ldots + \frac{1}{(5n-4)(5n+1)}$

$=$ $\frac{1}{(5 \times 1 - 4)(5 \times 1 + 1)} + \frac{1}{(5 \times 2 - 4)(5 \times 2 + 1)} + \frac{1}{(5 \times 3 - 4)(5 \times 3 + 1)}$ $+ \ldots$

$T_n =$ $\frac{1}{(5n-4)(5n+1)}$

$S_n = \sum \limits_{k=1}^n$ $\frac{1}{(5k-4)(5k+1)}$

$=$ $\frac{1}{5}$ $\Big[ \sum \limits_{k=1}^n \Big($ $\frac{1}{5k-4}$ $-$ $\frac{1}{5k+1}$ $\Big) \Big]$

$=$ $\frac{1}{5}$ $\Big[ \sum \limits_{k=1}^n$ $\frac{1}{5k-4}$ $- \sum \limits_{k=1}^n$ $\frac{1}{5k+1}$ $\Big]$

$=$ $\frac{1}{5}$ $\Big[ \Big( 1 +$ $\frac{1}{6} + \frac{1}{11} + \ldots + \frac{1}{5n-4}$ $\Big) - \Big($ $\frac{1}{6} + \frac{1}{11} + \ldots + \frac{1}{5n-4} + \frac{1}{5n+1}$ $\Big) \Big]$

$=$ $\frac{1}{5}$ $\Big[ 1 -$ $\frac{1}{5n+1}$ $\Big]$

$=$ $\frac{n}{5n+1}$