sum_(i=1)^n (4i^2(i-1))/n^4 = ((n+1))/(3n^3) ( 3n^2-n-2 )

sum_(i=1)^n (4i^2(i-1))/n^4 = ((n+1))/(3n^3) ( 3n^2-n-2 )

Let S_n = sum_(i=1)^n (4i^2(i-1))/n^4
:. S_n = 4/n^4sum_(i=1)^n (i^3-i^2)
:. S_n = 4/n^4{sum_(i=1)^n i^3 – sum_(i=1)^n i^2 }

And using the standard results:
sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) ” ; ” sum_(r=1)^n r^3 = 1/4n^2(n+1)^2
We have;

S_n = 4/n^4{ 1/4n^2(n+1)^2 – 1/6n(n+1)(2n+1)}
:. S_n = 4/n^4 ( (n(n+1))/12){ 3n(n+1) – 2(2n+1) }
:. S_n = ((n+1))/(3n^3) { 3n^2+3n-4n-2 }
:. S_n = ((n+1))/(3n^3) ( 3n^2-n-2 )

And this has been calculated using Excel for n=10, 100, 1000, 10000

What happens as n rarr oo?
[ NB As an additional task we could possibly conclude that as n rarr oo then S_n rarr 1; This is probably the conclusion of this question]
Now, S_n = ((n+1))/(3n^3) ( 3n^2-n-2 )

:. S_n = 1/(3n^3)( 3n^3-n^2-2n + 3n^2-n-2 )
:. S_n = 1/(3n^3)( 3n^3+2n^2 -3n-2)
:. S_n = 1/3( 3+2/n -3/n^2-2/n^3)

And so,

lim_(n rarr oo) S_n = lim_(n rarr oo) 1/3( 3+2/n -3/n^2-2/n^3)
:. lim_(n rarr oo) S_n = 1/3( 3+0 -0-0)
:. lim_(n rarr oo) S_n = 1

Which confirms our assumption!