If S_1 = sum_{r=1}^{n} r,S_2 = sum_{r=1}^{n} | vedclass

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(C) We know the formulas for the sums of powers of the first $n$ natural numbers: $S_1 = \frac{n(n+1)}{2}$,$S_2 = \frac{n(n+1)(2n+1)}{6}$,$S_3 = \frac

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$\frac{9}{32}$

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(C) We know the formulas for the sums of powers of the first $n$ natural numbers: $S_1 = \frac{n(n+1)}{2}$,$S_2 = \frac{n(n+1)(2n+1)}{6}$,$S_3 = \frac{n^2(n+1)^2}{4}$. Given the expression $L = \lim_{n \rightarrow \infty} \frac{S_1(1 + \frac{S_3}{4})}{S_2^2}$. Substituting the formulas: $L = \lim_{n \rightarrow \infty} \frac{\frac{n(n+1)}{2} (1 + \frac{n^2(n+1)^2}{16})}{\frac{n^2(n+1)^2(2n+1)^2}{36}}$. Simplifying the expression: $L = \lim_{n \rightarrow \infty} \frac{n(n+1)}{2} \cdot \frac{16 + n^2(n+1)^2}{16} \cdot \frac{36}{n^2(n+1)^2(2n+1)^2}$. $L = \lim_{n \rightarrow \infty} \frac{18}{16} \cdot \frac{n(n+1) [16 + n^2(n+1)^2]}{n^2(n+1)^2(2n+1)^2}$. $L = \frac{9}{8} \lim_{n \rightarrow \infty} \frac{16 + n^2(n+1)^2}{n(n+1)(2n+1)^2}$. Dividing the numerator and denominator by $n^4$: $L = \frac{9}{8} \lim_{n}$ ${\rightarrow \infty} \frac{\frac{16}{n^4} + (1 + \frac{1}{n})^2}{(1 + \frac{1}{n}) (2 + \frac{1}{n})^2} = \frac{9}{8} \cdot \frac{0 + 1}{1 \cdot 4} = \frac{9}{32}$.

If $S_1 = \sum_{r=1}^{n} r$,$S_2 = \sum_{r=1}^{n} r^2$,and $S_3 = \sum_{r=1}^{n} r^3$,then the value of $\lim_{n \rightarrow \infty} \frac{S_1(1 + \frac{S_3}{4})}{S_2^2}$ is

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