Find the sum of the following series to n terms:

Question 1: 3+5+9+15+23 + \ldots 

Answer:

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)

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Question 2: 2+5+10+17+26+  \ldots  

Answer:

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}

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Question 3: 1+3+7+13+21+ \ldots  

Answer:

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)

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Question 4: 3+7+14+24+37+ \ldots  

Answer:

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)

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Question 5: 1+3+6+10+15 + \ldots 

Answer:

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} 

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Question 6: 1+4+13+40+121 + \ldots 

Answer:

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}

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Question 7: 4+6+9+13+18+ \ldots 

Answer:

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)

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Question 8: 2+4+7+11+16 + \ldots 

Answer:

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)

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

Answer:

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}

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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)}

Answer:

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}