Generalized term be r(r + 1) (r + 4)

1^st term = (1)((1) + 1) ((1) + 4)

2^nd term =(2)((2) + 1) ((2) + 4)

And so on

nth term = n(n + 1) (n + 4) = n³ + 5n² + 4n

Summation = 1^st term + 2^nd term + …… + nth term

=(1)((1) + 1) ((1) + 4) + (2)((2) + 1) ((2) + 4)……… + n³ + 5n² + 4n ……(1)

We know,

Thus

From (1) we have

Summation =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus

(2)

We know,

Thus substituting the above values in (2)

Summation