A) 7820
B) 7830
C) 7520
D) 7510
Correct Answer: A
Solution :
\[{{T}_{n}}=\frac{(3+(n-1)\times 3)({{1}^{2}}+{{2}^{2}}+...+{{n}^{2}})}{(2n+1)}\] |
\[{{T}_{n}}=\frac{3.\frac{{{n}^{2}}(n+1)(2n+1)}{6}}{2n+1}+\frac{{{n}^{2}}(n+1)}{2}\] |
\[{{S}_{15}}=\,\frac{1}{2}\sum\limits_{n\,\,=\,\,1}^{15}{({{n}^{3}}+{{n}^{2}})}\] |
\[=\,\frac{1}{2}\left[ {{\left( \frac{15(15+1)}{2} \right)}^{2}}+\frac{15\times 16\times 31}{6} \right]\] |
= 7820. |
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