Among (S1): _ n → ∞ 1 n ^ 2 ( 2 + 4 + 6 + + 2n ) = 1 (S2): _ n → ∞...

${S}_{1}: \lim_{n \to \infty}\dfrac{1}{{n}^{2}} [ 2 + 4 + 6 + \ldots + 2n ]$

$\lim_{n \to \infty}2\dfrac{n ( n + 1 )}{2{n}^{2}} = 1$

${S}_{2}: \lim_{n \to \infty}\dfrac{1}{{n}^{16}}\left({1}^{15} + {2}^{15} + {3}^{15} + \ldots + {n}^{15}\right)$

This is limit of sum.

We know that $\lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}f\left(\dfrac{r}{n}\right)\dfrac{1}{n} = {∫}_{0}^{1}f\left(x\right) d x$

$∵ \lim_{n \to \infty}\dfrac{{∑}_{r = 1}^{n}{r}^{k}}{{n}^{k + 1}} = \lim_{n \to \infty}\dfrac{{∑}_{r = 1}^{n}{\left(\dfrac{r}{n}\right)}^{k}}{n}$

$= {∫}_{0}^{1}{x}^{k} d x = \dfrac{1}{k + 1}$

Here $k = 15$

$∴ \lim_{n \to \infty}\dfrac{{∑}_{r = 1}^{n}{r}^{15}}{{n}^{16}} = \dfrac{1}{16}$

$∴$ Both $S1$ and $S2$ are correct

Hence this is the required option.