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  • If  <img alt="\mathrm{\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\ldots+(n k+n)]}" src="/latex-image/?%5Cmathrm%7B%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7D%20%

If  \mathrm{\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\ldots+(n k+n)]}\mathrm{=33 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot\left[1^{k}+2^{k}+3^{k}+\ldots+n^{k}\right]},

then the integral value of \mathrm{k} is equal to _______.

Option: 1

5


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n k+1}[n k \cdot n+1+2+\cdots+n]}

\mathrm{=\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}} \cdot\left[n^{2} k+\frac{(n(n+1)}{2}\right]}

\mathrm{\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1} \cdot n^{2}\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)}{n k+1}}

\mathrm{\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)}

\mathrm{\Rightarrow\left(k+\frac{1}{2}\right)}

RHS

\mathrm{\Rightarrow \lim _{n \rightarrow \infty} \frac{1}{n k+1}\left(1^{k}+2^{k}+\cdots \cdot+h^{k}\right)=\frac{1}{k+1}}

LHS=RHS

\mathrm{\Rightarrow k+\frac{1}{2}=33 \cdot \frac{1}{k+1}} \\

\mathrm{\Rightarrow(2 k+1)(k+1)=66} \\

\mathrm{\Rightarrow(k-5)(2 k+13)=0} \\

\mathrm{\Rightarrow k=5 \text { or } \frac{13}{2}}

Hence answer is 5

Posted by

chirag

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