Find the sum to n terms of the series 1 ^ 2 + ( 1 ^ 2 + 2 ^ 2 ) + ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 ) +

Given: 1^2 + (1^2 + 2^2) +(1^2 + 2^2 + 3^2)+....... to n terms a^n = (1^2 + 2^2 + 3^2 + ....... n^2) = n ( n + 1 ) ( 2 n + 1 ) 6 = 1 6 ( 2 n ^ 3 + 3 n ^ 2 + n ) S _n = _ k = 1 ^n a_k = _ k = 1 ^n 1 6 ( 2 k^3 + 3 k^2 + k ) = 1 6 ( 2.1 ^ 3 + 3.1 ^ 2 + 1 ) + 1 6 ( 2.2 ^ 3 + 3.2 ^ 2 + 2 ) + + 1 6 ( 2 . n ^ 3 + 3 n ^ 2 + n ) = 1 6 [ 2 ( 1 ^ 3 + 2 ^ 3 + + n ^ 3 ) + 3 ( 1 ^ 2 + 2 ^ 2 + + n ^ 2 ) + ( 1 + 2 + 3 + + n ) ] = 1 6 [ 2 n ( n + 1 ) 2 ^ 2 + 3 n ( n + 1 ) ( 2 n + 1 ) 6 + n ( n + 1 ) 2 ] = n ( n + 1 ) 12 [ n ( n + 1 ) + 2 n + 1 + 1 1 ] = n ( n + 1 ) 12 ( n ^ 2 + 3 n + 2 ) = n ( n + 1 ) ( n + 1 ) ( n + 2 ) 12 = n ( n + 1 ) ^ 2 ( n + 2 ) 12

Find the sum to n terms of the series 1 ^ 2 + ( 1 ^ 2 + 2 ^ 2 ) +…

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Question

Find the sum to n terms of the series $1 ^ { 2 } + \left( 1 ^ { 2 } + 2 ^ { 2 } \right) + \left( 1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } \right) + \ldots \ldots$

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Answer

Given: $1^2 + (1^2 + 2^2) +(1^2 + 2^2 + 3^2)+$....... to $n$ terms$\therefore a^n = (1^2 + 2^2 + 3^2 + ....... n^2)$
= $\frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }$
= $\frac { 1 } { 6 } \left( 2 n ^ { 3 } + 3 n ^ { 2 } + n \right)$
${\therefore} \ {{\text{S}}_n} = \sum\limits_{k = 1}^n {{a_k}} = \sum\limits_{k = 1}^n {\frac{1}{6}} \left( {2{k^3} + 3{k^2} + k} \right)$
$=\frac { 1 } { 6 } \left( 2.1 ^ { 3 } + 3.1 ^ { 2 } + 1 \right) + \frac { 1 } { 6 } \left( 2.2 ^ { 3 } + 3.2 ^ { 2 } + 2 \right) + \ldots \ldots + \frac { 1 } { 6 } \left( 2 . n ^ { 3 } + 3 n ^ { 2 } + n \right)$
$=\frac { 1 } { 6 } \left[ 2 \left( 1 ^ { 3 } + 2 ^ { 3 } + \ldots \ldots + n ^ { 3 } \right) + 3 \left( 1 ^ { 2 } + 2 ^ { 2 } + \ldots \ldots + n ^ { 2 } \right) + ( 1 + 2 + 3 + \ldots \ldots + n ) \right]$
$= \frac { 1 } { 6 } \left[ 2 \left\{ \frac { n ( n + 1 ) } { 2 } \right\} ^ { 2 } + \frac { 3 n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \frac { n ( n + 1 ) } { 2 } \right]$
$= \frac { n ( n + 1 ) } { 12 } \left[ \frac { n ( n + 1 ) + 2 n + 1 + 1 } { 1 } \right]$
$= \frac { n ( n + 1 ) } { 12 } \left( n ^ { 2 } + 3 n + 2 \right)$
$= \frac { n ( n + 1 ) ( n + 1 ) ( n + 2 ) } { 12 }$
$= \frac { n ( n + 1 ) ^ { 2 } ( n + 2 ) } { 12 }$

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