Verified answer

1. ∑(2i/n)(2/n) = (4/n^2)[1 + 2 + 3 + ....]

[i=1, n] ∑(2i/n)(2/n) = (4/n^2)[n(n + 1)/2] = 2(n + 1)/n = 2 + 2/n

As n --> ∞, 2/n --> 0, so

lim n --> ∞ [i=1, n]∑(2i/n)(2/n) = 2

2. [i=1, n]∑ (1/n^3)(i - 1)^2 = (1/n^3)[0 + 1 + 4 + 9 + ...]

= (1/n^3)[n(n - 1)(2n - 1)/6] = (2n^3 - 3n^2 + n)/6n^3

lim n --> ∞ [i=1, n]∑ (1/n^3)(i - 1)^2 = 1/3

3. Looks like possible misprint as no i within sigma term.

Also presume expression below was meant, rather than ()^[2(2/n)]

∑ [(1+2i/n)^2]*(2/n) = (2/n^3) ∑ [(n + 2i)^2]

Let T = [i=1, n] ∑ [(2i + n)^2] and expand using

(2i + n)^2 = 4i^2 + 4ni + n^2

Let A = 4[i=1, n] ∑i^2 = (8n^3 + 12n^2 + 4n)/6

Let B = 4n[i=1, n] ∑i = 4n(n^2 + n)/2 = (12n^3 + 12n^2)/6

Let C = [i=1, n] ∑n^2 = n^3 = 6n^3/6

T = (13n^3 +12n^2 + 2n)/3

Now multiply this by the (2/n^3) factor

∑ [(1+2i/n)^2]*(2/n) = 26/3 + 8/n + 4/3n^2

The limit as n --> ∞ of this is just 26/3

Regards - Ian