Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).

Given,

a_n = n (n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n

Now, the sum of n terms of the series can be expressed as

\( S_n= \displaystyle\sum_{k=1}^{n}a_k\)

The sum of n terms of the given series is

\(s_n = 4\displaystyle\sum_{k=1}^{n}k^3+5\displaystyle\sum_{k=1}^{n}k^2 +4\displaystyle\sum_{k=1}^{n}k\)

= \( \dfrac{n^2(n+1)^2}{4}+\dfrac{5n(n+1)(2n+1)}{6}\)

\( +\dfrac{4n(n+1)}{2}\)

\(\dfrac{n(n+1)}{2}+[\dfrac{n(n+1)}{2} +\)

= \( \dfrac{5n(n+1)}{3}+4]\)

= \( \dfrac{n(n+1)}{2}\)

= \( [\dfrac{3n^2+3n+20n+10+14}{6} ]\)

\( \dfrac{n(n+1)}{2}= [\dfrac{3n^2+23n+34}{6} ]\)

= \(\dfrac{n(n+1)(3n^2+23n+34)}{12} \)

Therefore, the sum of n^th terms of the series whose n^th term is given by

= \(\dfrac{n(n+1)(3n^2+23n+34)}{12} \)