Solve the following


3 × 12 + 5 ×22 + 7 × 32 + ...


Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=(2 n+1) n^{2}=2 n^{3}+n^{2}$

Now, let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:

$S_{n}=\sum_{k=1}^{n} T_{k}$

$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(2 k^{3}+k^{2}\right)$

$\Rightarrow S_{n}=2 \sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n} k^{2}$

$\Rightarrow S_{n}=\left[\frac{2 n^{2}(n+1)^{2}}{4}+\frac{n(n+1)(2 n+1)}{6}\right]$

$\Rightarrow S_{n}=\left[\frac{n^{2}(n+1)^{2}}{2}+\frac{n(n+1)(2 n+1)}{6}\right]$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left[n(n+1)+\frac{2 n+1}{3}\right]$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{3 n^{2}+3 n+2 n+1}{3}\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{3 n^{2}+5 n+1}{3}\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{6}\left(3 n^{2}+5 n+1\right)$





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