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Among
$$(S1) : \lim_{n \to \infty} \frac{1}{n^2}(2 + 4 + 6 + \ldots + 2n) = 1$$
$$(S2) : \lim_{n \to \infty} \frac{1}{n^{16}}(1^{15} + 2^{15} + 3^{15} + \ldots + n^{15}) = \frac{1}{16}$$
Given:
$$(S1): \lim_{n\to\infty}\frac{2+4+6+...+2n}{n^2} = \lim_{n\to\infty}\frac{n(n+1)}{n^2} = 1$$. TRUE ✓
$$(S2): \lim_{n\to\infty}\frac{1^{15}+2^{15}+...+n^{15}}{n^{16}}$$.
By Riemann sum interpretation: $$= \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^{15} = \int_0^1 x^{15}dx = \frac{1}{16}$$. TRUE ✓
Both are true. The answer is Option A: Both (S1) and (S2) are true.
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