I have done it a few times, but I can't seem to get the right answer:

Limit n-> infinity : nE(i=1)_(1+(2i)/n)^2*(2/n)

here is my best attempt:

(limit in front for all of this)
1. move the 2/n over to get: 2/n*(nE(i=1)_(1+(2i)/n)^2)
2. foil to get: 2/n*(nE(i=1)_(1+4i/n+4i^2/n^2)
3. break it put into further sigma: 2/n*(nE(i=1)_1+nE(i=1)_4i/n+nE(i=1)_4i^2/n^2)
4. move over shit: 2/n*(nE(i=1)_1+(4/n)*nE(i=1)_i+(4/n^2)*nE(i=1)_i^2)
5. simplify the sigmas: 2/n*(n+(4/n)*((n)(n+1)/2)+(4/n^2)*((n)(n+1)(2n+2))/6))
6. simplifying: 2/n*(n+2(n+1)+(2*(n+1)(2n+2)/(3*n)))

at this point I am not quite sure what to do to be honest, I am not sure how to simplify the 3rd term more tied in with the front 2/n

any ideas?