# The sum of the following series $1+6+\frac{9(1^{2}+2^{2}+3^{2})}{7}+\frac{12(1^{2}+2^{2}+3^{2}+4^{2})}{9}+\frac{15(1^{2}+2^{2}+3^{2}+.....+5^{2}}{11})+.......$up to 15 terms, is:Option 1)  7830Option 2)  7820Option 3)  7510Option 4)  7520

Summation of series of natural numbers -

$\sum_{k=1}^{n}K= \frac{1}{2}n\left ( n+1 \right )$

- wherein

Sum of first n natural numbers

$1+2+3+4+------+n= \frac{n(n+1)}{2}$

Summation of series of natural numbers -

$\sum_{k=1}^{n}K^{2}= \frac{1}{6}n\left ( n+1 \right )\left ( 2n+1 \right )$

- wherein

Sum of  squares of first n natural numbers

$1^{2}+2^{2}+3^{2}+4^{2}+------+n^{2}= \frac{n(n+1)\left ( 2n+1 \right )}{6}$

Summation of series of natural numbers -

$\sum_{k=1}^{n}K^{3}=\left ( \sum_{k=1}^{n}K \right )^{2}$

- wherein

$1^{3}+2^{3}+3^{3}+-------n^{3}\\= \left ( 1+2+3+-----+n \right )^{2}$

The general term of the sequence will be

$T_{n} = \frac{(3+(n-1)\times 3)(1^2 +2^2 +3^2+...+n^2)}{(2n+1)}$

$T_{n} = \frac{3\cdot \frac{n(n+1)(2n+1)}{6}}{(2n+1)} = \frac{n^2(n+1)}{2}$

$S_15 = \sum_{n=1}^{15}T_n = \frac{1}{2}\sum_{n=1}^{15}(n^3 + n^2) \\ = \frac{1}{2}\left[\left(\frac{15(15+1)}{2} \right )^2 + \frac{15\times 16\times 31}{6} \right ] = 7820$

Option 1)

7830

Option 2)

7820

Option 3)

7510

Option 4)

7520

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