Question

Find the sum of the series whose nth term is:

n (n + 1) (n + 4)

Answer 1

Let \[T_n\] be the nth term of the given series.
Thus, we have:

\[T_n = n\left( n + 1 \right)\left( n + 4 \right) = \left( n^2 + n \right)\left( n + 4 \right) = n^3 + 5 n^2 + 4n\]

Let

\[S_n\] be the sum of n terms of the given series.
Now, 

\[S_n = \sum^n_{k = 1} T_k\]

\[\Rightarrow S_n = \sum^n_{k = 1} \left( k^3 + 5 k^2 + 4k \right)\]

\[ \Rightarrow S_n = \sum^n_{k = 1} k^3 + {5\sum}^n_{k = 1} k^2 + 4 \sum^n_{k = 1} k\]

\[ \Rightarrow S_n = \frac{n^2 \left( n + 1 \right)^2}{4} + \frac{5n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + \frac{4n\left( n + 1 \right)}{2}\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left[ \frac{n\left( n + 1 \right)}{2} + \frac{5\left( 2n + 1 \right)}{3} + 4 \right]\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{12}\left[ 3n\left( n + 1 \right) + 10\left( 2n + 1 \right) + 24 \right]\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{12}\left( 3 n^2 + 23n + 34 \right)\]