A) \[\frac{1}{6}({{n}^{2}}+3n+8)\]
B) \[\frac{n}{6}({{n}^{2}}+3n+8)\]
C) \[\frac{1}{6}({{n}^{2}}-3n+8)\]
D) \[\frac{n}{6}({{n}^{2}}-3n+8)\]
Correct Answer: B
Solution :
[b] We have \[S=2+4+7+11+16+....+{{T}_{n}}\] Again \[S=2+4+7+11+16+....+{{T}_{n-1}}+{{T}_{n}}\] Subtracting, we get \[0=2+\{2+3+4+5+.....+({{T}_{n}}-{{T}_{n-1}})\}-{{T}_{n}}\] \[{{T}_{n}}=2+\frac{1}{2}(n-1)\,(4+\{n-2\}1)=\frac{1}{2}({{n}^{2}}+n+2)\] Now \[S=\Sigma {{T}_{n}}=\frac{1}{2}\Sigma ({{n}^{2}}+n+2)=\frac{1}{2}(\Sigma {{n}^{2}}+\Sigma n+2\Sigma 1)\] \[=\frac{1}{2}\left\{ \frac{1}{6}n(n+1)(2n+1)+\frac{1}{2}n(n+1)+2n \right\}\] \[=\frac{n}{12}\{(n+1)(2n+1+3)+12\}\] \[=\frac{n}{6}\{(n+1)(n+2)+6\}=\frac{n}{6}({{n}^{2}}+3n+8)\]
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