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