Question

Find the sum of n terms of the series whose nth terms is n²(n+4)=n³+4n²

Answer 1

Answer:

nth term  is n4−n2

Sum of n terms is Σ(n4−n2)=Σn4−Σn2

now we know Σn4=30n(n+1)(2n+1)(3n2+3n−1)

and Σn2=6n(n+1)(2n+1)

So,  Σ(n4−n2)=30n(n+1)(2n+1)(3n2+3n−1)−6n(n+1)(2n+1)

⇒Σ(n4−n2)=10n(n+1)(2n+1)(n2+n−2)

⇒Σ(n4−n2)=10n(n+1)(2n+1)(n+2)(n−1)