The sum of the follwing series

Question:

The sum of the follwing series

$1+6+\frac{9\left(1^{2}+2^{2}+3^{2}\right)}{7}+\frac{12\left(1^{2}+2^{2}+3^{2}+4^{2}\right)}{9}$

$+\frac{15\left(1^{2}+2^{2}+\ldots+5^{2}\right)}{11}+\ldots$ up to 15 terms, is:

  1. 7820

  2. 7830

  3. 7520

  4. 7510


Correct Option: 1

Solution:

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

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

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

$=7820$

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