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  • Among <img alt="\mathrm{(S 1)}: \lim _{n \rightarrow \infty} \frac{1}{n^2}(2+4+6+\ldots . .+2 n)=1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28S%201%29%7D%3A%20%5

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\mathrm{(S 1)}: \lim _{n \rightarrow \infty} \frac{1}{n^2}(2+4+6+\ldots . .+2 n)=1

\mathrm{(S2)}: \lim _{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots . .+n^{15}\right)=\frac{1}{16}

Option: 1

Only (S1) is true
 


Option: 2

Both (S1) and (S2) are true
 


Option: 3

Both (S1) and (S2) are false


Option: 4

Only (S2) is true
 


Answers (1)

best_answer

\begin{aligned} & S_1: \lim _{n \rightarrow \infty} \frac{n(n+1)}{n^2}=1 \Rightarrow \text { True } \\ & S_2: \lim _{n \rightarrow \infty} \frac{1}{n^{16}}\left(\sum r^{15}\right)=\lim _{n \rightarrow \infty} \frac{1}{n} \sum\left(\frac{r}{n}\right)^{15} \\ \end{aligned}

\begin{aligned} & =\int_0^1 x^{15} d x=\frac{1}{16} \Rightarrow \text { True } \end{aligned}

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Riya

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