Q91 / 155 · WBJEE 2024 · Mathematics

$\lim _{x \rightarrow \infty} \frac{1}{n^{k+1}}\left[2^k+4^k+6^k+\ldots .+(2 n)^k\right]=$

  • A
    $\frac{2^k}{k}$
  • B
    $\frac{2^{\mathrm{k}+1}}{\mathrm{k}+1}$
  • Correct answer
    C
    $\frac{2^k}{\mathrm{k}+1}$
  • D
    $\frac{2^{\mathrm{k}}}{\mathrm{k}-1}$

Solution

Hint: $\lim _{\mathrm{x} \rightarrow \infty} \frac{1}{\mathrm{n}} \sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\frac{2 \mathrm{r}}{\mathrm{n}}\right)^{\mathrm{k}}=\int_0^1(2 \mathrm{x})^{\mathrm{k}} \mathrm{dx}=\frac{2^{\mathrm{k}}}{\mathrm{k}+1}$

Hence, the correct answer is option: 'C'.

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