Answer: (i) \(S_{2n}=-n(2n+1)\)

(ii) \(\displaystyle \lim_{n\to\infty}\frac{S_{2n}}{n^2}=-2\), and \(\displaystyle \lim_{n\to\infty}\frac{S_{2n+1}}{n^2}=2\).

We are given \(S_n=\sum_{r=1}^n (-1)^{r-1}r^2\).

(i) Consider \(S_{2n}\):

\(S_{2n}=1^2-2^2+3^2-4^2+\cdots+(2n-1)^2-(2n)^2\).

Group the terms in pairs:

\(S_{2n}=(1^2-2^2)+(3^2-4^2)+\cdots+((2n-1)^2-(2n)^2)\).

For the \(k\)th pair,

\((2k-1)^2-(2k)^2=(4k^2-4k+1)-4k^2=1-4k\).

So

\(S_{2n}=\sum_{k=1}^n (1-4k)=n-4\sum_{k=1}^n k\).

Using \(\sum_{k=1}^n k=\frac{n(n+1)}{2}\),

\(S_{2n}=n-4\cdot\frac{n(n+1)}{2}=n-2n(n+1)=n-2n^2-2n=-2n^2-n\).

Hence

\(S_{2n}=-n(2n+1)\).

(ii) From part (i),

\(\frac{S_{2n}}{n^2}=\frac{-n(2n+1)}{n^2}=-\left(2+\frac1n\right)\).

Therefore

\(\lim_{n\to\infty}\frac{S_{2n}}{n^2}=-2\).

Now find \(S_{2n+1}\). Since

\(S_{2n+1}=S_{2n}+(2n+1)^2\),

we have

\(S_{2n+1}=-n(2n+1)+(2n+1)^2\).

Factor out \((2n+1)\):

\(S_{2n+1}=(2n+1)(-n+2n+1)=(2n+1)(n+1)\).

So

\(\frac{S_{2n+1}}{n^2}=\frac{(2n+1)(n+1)}{n^2}=(2+\frac1n)(1+\frac1n)\).

Taking the limit gives

\(\lim_{n\to\infty}\frac{S_{2n+1}}{n^2}=2\).