Find the sum of the following series to \displaystyle n terms:

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

Answer:

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

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

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

Subtracting ii) from i) we get

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\displaystyle \\

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

Answer:

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

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

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

Subtracting ii) from i) we get

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

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

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

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

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

\displaystyle \Rightarrow T_n = n^2+1

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

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

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

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

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

\displaystyle \\

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

Answer:

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

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

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

Subtracting ii) from i) we get

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\displaystyle \\

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

Answer:

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

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

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

Subtracting ii) from i) we get

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

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

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

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

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

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

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

\displaystyle \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

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

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

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

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

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

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

\displaystyle \\

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

Answer:

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

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

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

Subtracting ii) from i) we get

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

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

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

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

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

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

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

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

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

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

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

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

\displaystyle \\

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

Answer:

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

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

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

Subtracting ii) from i) we get

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

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

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

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

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

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

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

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

\displaystyle \\

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

Answer:

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

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

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

Subtracting ii) from i) we get

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

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

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

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

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

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

\displaystyle \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

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

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

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

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

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

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

\displaystyle \\

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

Answer:

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

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

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

Subtracting ii) from i) we get

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

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

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

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

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

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

\displaystyle \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

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

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

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

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

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

\displaystyle \\

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

Answer:

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

\displaystyle = \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

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

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

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

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

\displaystyle = \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]

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

\displaystyle = \frac{n}{3n+1}  

\displaystyle \\

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

Answer:

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

\displaystyle = \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

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

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

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

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

\displaystyle = \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]

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

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