We have
an = n(n + 1) (n + 4) = (n2 + n) (n + 4) = n3+4n2+n2 + 4n = n3+5n2 + 4n
Hence, the sum to n terms is
Sn =

n

Σ

k=1

ak =

n

Σ

k=1

(k3+5k2+4k) =

n

Σ

k=1

k3+5

n

Σ

k=1

k2+4

n

Σ

k=1

k
= [

n(n+1)

2

]2 + 5[

n(n+1)(2n+1)

6

] + 4[

n(n+1)

2

]


=

n(n+1)

2

[

3n2+23n+34

6

] =

n(n+1)(n+2)(3n+17)

12