The sum of first 9 terms of the series.

Question

$${{{1^3}} \over 1} + {{{1^3} + {2^3}} \over {1 + 3}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 3 + 5}} + ......$$

Accepted Answer

${n^{th}}$ term of series

$$ = {{\left[ {{{n\left( {n + 1} \right)} \over 2}} \right]} \over {{n^2}}} = {1 \over 4}{\left( {n + 1} \right)^2}$$

Sum of $n$ term $$ = \sum {{1 \over 4}} {\left( {n + 1} \right)^2}$$

$$ = {1 \over 4}\left[ {\sum {n{}^2}  + 2\sum n  + n} \right]$$

$$ = {1 \over 4}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6} + {{2n\left( {n + 1} \right)} \over 2} + n} \right]$$

Sum of $9$ terms

$$ = {1 \over 4}\left[ {{{9 \times 10 \times 19} \over 6} + {{18 \times 10} \over 2} + 9} \right] = {{384} \over 4} = 96$$

Solutions · 1