If lim_{n rightarrow infty} frac{(n+1)^{k-1}} | vedclass

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(B) Let $LHS = \lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}} [nk \cdot n + \frac{n(n+1)}{2}]$$= \lim_{n \rightarrow \infty} \frac{(n+1)^{k-1

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Let $LHS = \lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}} [nk \cdot n + \frac{n(n+1)}{2}]$$= \lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}} \cdot n^2 [k + \frac{1 + \frac{1}{n}}{2}]$$= \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^{k-1} \cdot (k + \frac{1 + \frac{1}{n}}{2}) = k + \frac{1}{2}$.Let $RHS = \lim_{n \rightarrow \infty} \frac{1}{n^{k+1}} \sum_{i=1}^{n} i^k = \int_{0}^{1} x^k dx = \frac{1}{k+1}$.Given $LHS = 33 \cdot RHS$,so $k + \frac{1}{2} = 33 \cdot \frac{1}{k+1}$.$(2k + 1)(k + 1) = 66$.$2k^2 + 3k + 1 = 66 \implies 2k^2 + 3k - 65 = 0$.$(2k + 13)(k - 5) = 0$.Since $k$ is an integral value,$k = 5$.

If $\lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+\ldots+(nk+n)] = 33 \cdot \lim_{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot [1^k + 2^k + 3^k + \ldots + n^k]$,then the integral value of $k$ is equal to $....$

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