Among S 1 : lim n → ∞ 1 n 2 ( 2 + 4 + 6 + … + 2 n ) = 1 S 2 : lim n → ∞ 1 n 16…

$S_1:\underset{n\to \infty }{\lim }\frac{1}{n^2}[2+4+6+\dots +2n]$

$\underset{n\to \infty }{\lim }2\frac{n(n+1)}{2n^2}=1$

$S_2:\underset{n\to \infty }{\lim }\frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\dots +n^{15}\right)$

This is limit of sum.

We know that $\underset{n\to \infty }{\lim }\sum _{r=1}^{n}f\left(\frac{r}{n}\right)\frac{1}{n}={\int }_0^{1}f\left(x\right)dx$

$\because \underset{n\to \infty }{\lim }\frac{{\sum }_{r=1}^n{r}^k}{n^{k+1}}=\underset{n\to \infty }{\lim }\frac{{\sum }_{r=1}^n{\left({\displaystyle \frac{r}{n}}\right)}^k}{n}$

$={\int }_0^1{x}^{k}dx=\frac{1}{k+1}$

Here $k=15$

$\therefore \underset{n\to \infty }{\lim }\frac{{\sum }_{r=1}^n{r}^{15}}{n^{16}}=\frac{1}{16}$

$\therefore$ Both $S1$ and $S2$ are correct

Hence this is the required option.