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

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Question

(B) ધારો કે $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)^

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) ધારો કે $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}$.ધારો કે $RHS = \lim_{n \rightarrow \infty} \frac{1}{n^{k+1}} \sum_{i=1}^{n} i^k = \int_{0}^{1} x^k dx = \frac{1}{k+1}$.આપેલ છે કે $LHS = 33 \cdot RHS$,તેથી $k + \frac{1}{2} = 33 \cdot \frac{1}{k+1}$.$(2k + 1)(k + 1) = 66$.$2k^2 + 3k + 1 = 66 \implies 2k^2 + 3k - 65 = 0$.$(2k + 13)(k - 5) = 0$.કારણ કે $k$ એ પૂર્ણાંક કિંમત છે,તેથી $k = 5$.

જો $\lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+\ldots+(nk+n)] = 33 \cdot \lim_{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot [1^k + 2^k + 3^k + \ldots + n^k]$ હોય,તો $k$ ની પૂર્ણાંક કિંમત $....$ છે.

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