How to solve the limit of this sum?

For this sort of problem, you just multiply out the cube to get something in the form:

Σ(A + Bi + Ci² + Di³) = nA + BΣi + CΣi² + DΣi³

...then use the given closed forms for each of those sums on the right. By inspection, I can see that A will be (4 + 5)(2/n) = 18/n, so nA = 18. No problem with a limit there. It's just 18 for every value of n>0.

Also D must be 4(2/n)³(1/n) so:

D Σi³ = (32 / n^4) n² (n+1)² / 4

= 8 (n + 1)² / n² = 8 (1 + 1/n)²

The 1/n term vanishes as n-->oo, so that term has 8 as a limit.

I'll let you work out the other two terms.