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$2+4+7+11+16+\ldots \ldots$ to $n$ terms $=$
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Verified Answer
The correct answer is:
$\frac{n}{6}\left(n^2+3 n+8\right)$
We have $S=2+4+7+11+16+\ldots .+T_n$
Again $S=2+4+7+11+\ldots \ldots+T_{n-1}+T_n$
Subtracting, we get
$0=2+\left\{2+3+4+5+\ldots . .\left(T_n-T_{n-1}\right)\right\}-T_n$
$T_n=2+\frac{1}{2}(n-1)(4+\{n-2) 1\}=\frac{1}{2}\left(n^2+n+2\right)$
Now, $S=\Sigma T_n=\frac{1}{2} \Sigma\left(n^2+n+2\right)=\frac{1}{2}\left(\Sigma n^2+\Sigma n+2 \Sigma 1\right)$
$=\frac{1}{2}\left\{\frac{1}{6} n(n+1)(2 n+1)+\frac{1}{2} n(n+1)+2 n\right\}$
$=\frac{n}{12}\{(n+1)(2 n+1+3)+12\}$
$=\frac{n}{6}\{(n+1)(n+2)+6\}=\frac{n}{6}\left(n^2+3 n+8\right)$
Again $S=2+4+7+11+\ldots \ldots+T_{n-1}+T_n$
Subtracting, we get
$0=2+\left\{2+3+4+5+\ldots . .\left(T_n-T_{n-1}\right)\right\}-T_n$
$T_n=2+\frac{1}{2}(n-1)(4+\{n-2) 1\}=\frac{1}{2}\left(n^2+n+2\right)$
Now, $S=\Sigma T_n=\frac{1}{2} \Sigma\left(n^2+n+2\right)=\frac{1}{2}\left(\Sigma n^2+\Sigma n+2 \Sigma 1\right)$
$=\frac{1}{2}\left\{\frac{1}{6} n(n+1)(2 n+1)+\frac{1}{2} n(n+1)+2 n\right\}$
$=\frac{n}{12}\{(n+1)(2 n+1+3)+12\}$
$=\frac{n}{6}\{(n+1)(n+2)+6\}=\frac{n}{6}\left(n^2+3 n+8\right)$
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