Solve the following

Solution:

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

Thus, we have:

$T_{n}=n(n+1)(n+4)$

$=n\left(n^{2}+5 n+4\right)$

 

$=\left(n^{3}+5 n^{2}+4 n\right)$

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

Thus, we have:

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

$\Rightarrow S_{n}=\sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n} 5 k^{2}+\sum_{k=1}^{n} 4 k$

$\Rightarrow S_{n}=\sum_{k=1}^{n} k^{3}+5 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k$

$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{4}+\frac{5 n(n+1)(2 n+1)}{6}+\frac{4 n(n+1)}{2}$

$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{4}+\frac{5 n(n+1)(2 n+1)}{6}+2 n(n+1)$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{n(n+1)}{2}+\frac{5(2 n+1)}{3}+4\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{n^{2}+n}{2}+\frac{10 n+5}{3}+4\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{3 n^{2}+3 n+20 n+10+24}{6}\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{12}\left(3 n^{2}+23 n+34\right)$