The sum of the series 1+6+9(1^2+2^2+3^2)7+12(1^2+2^2+3^2+4^2)9+15(1^2+2^2++5^2)11+ up to 1…

Question

The sum of the series $1+6+\dfrac{9(1^2+2^2+3^2)}{7}+\dfrac{12(1^2+2^2+3^2+4^2)}{9}+\dfrac{15(1^2+2^2+\cdots+5^2)}{11}+\cdots$ up to 15 terms is:

Accepted Answer

$$1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9} + {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....\,15$$

$$ = {{3\left( {{1^2}} \right)} \over 3} + {{6\left( {{1^2} + {2^2}} \right)} \over 5} + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12} \over 9}\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right) + ......$$

$${T_r} = {{3r} \over {2r + 1}}\left( {{1^2} + {2^2} + .... + {r^2}} \right)$$

$${T_r} = {{3r} \over {2r + 1}}{{r\left( {r + 1} \right)\left( {2r + 1} \right)} \over 6} = {1 \over 2}{r^2}\left( {r + 1} \right)$$

Sum of $n$ terms $$ = \sum\limits_{r = 1}^n {{T_r}}  = {1 \over 2}\sum\limits_{r = 1}^n {\left( {{r^3} + {r^2}} \right)} $$

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

Sum upto 15 terms $$ \Rightarrow $$ then put $n$ = 15

$$ = {1 \over 2}\left( {{{{{\left( {15 \times 16} \right)}^2}} \over 4} + {{15 \times 16 \times 31} \over 6}} \right) = 7820$$

Solutions · 1