Q102 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n\left(k^2 x\right)=$

  • A
    $\mathbf{X}$
  • B
    $\frac{x}{2}$
  • Correct answer
    C
    $\frac{x}{3}$
  • D
    $\frac{x}{4}$

Solution

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n\left(k^2 x\right)$ $\begin{aligned} & =\lim _{n \rightarrow \infty} \frac{1}{n^3}\left(1^2+2^2+\ldots .+n^2\right) x \\ & =\lim _{n \rightarrow \infty} \frac{1}{n^3} \frac{n(n+1)(2 n+1)}{6} \times x \\ & \lim _{n \rightarrow \infty} \frac{x}{6}\left(2+\frac{3}{n}+\frac{1}{n^2}\right)=\frac{2 x}{6}=\frac{x}{3}\end{aligned}$

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

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