Among the following statements:(S1): lim _{n | vedclass

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(A) For $(S1)$: The sum of the first $n$ even numbers is $2(1+2+3+\ldots+n) = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Thus,$\lim _{n ightarrow \infty} \

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Both $(S1)$ and $(S2)$ are true

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(A) For $(S1)$: The sum of the first $n$ even numbers is $2(1+2+3+\ldots+n) = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Thus,$\lim _{n ightarrow \infty} \frac{n(n+1)}{n^2} = \lim _{n ightarrow \infty} \frac{n^2+n}{n^2} = \lim _{n ightarrow \infty} (1 + \frac{1}{n}) = 1$. So,$(S1)$ is true.For $(S2)$: We use the definition of a definite integral as the limit of a sum: $\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(\frac{r}{n}) = \int_{0}^{1} f(x) dx$.Here,$\lim _{n ightarrow \infty} \frac{1}{n^{16}} \sum_{r=1}^{n} r^{15} = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} (\frac{r}{n})^{15} = \int_{0}^{1} x^{15} dx$.Evaluating the integral: $\int_{0}^{1} x^{15} dx = [\frac{x^{16}}{16}]_{0}^{1} = \frac{1}{16}$. So,$(S2)$ is true.

Among the following statements:
$(S1): \lim _{n \rightarrow \infty} \frac{1}{n^2}(2+4+6+\ldots+2n)=1$
$(S2): \lim _{n \rightarrow \infty} \frac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+\ldots+n^{15})=\frac{1}{16}$

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Among the following statements:$(S1): \lim _{n \rightarrow \infty} \frac{1}{n^2}(2+4+6+\ldots+2n)=1$$(S2): \lim _{n \rightarrow \infty} \frac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+\ldots+n^{15})=\frac{1}{16}$