TS EAMCET · Maths · Sequences and Series
The value of \(\lim _{\substack{n \rightarrow \infty}} \frac{1}{n^3} \sum_{k=1}^n\left(k^2 x\right)\) is
- A \(x\)
- B \(\frac{x}{2}\)
- C \(\frac{x}{3}\)
- D \(\frac{x}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{x}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n\left(k^2\right. & x) \\ & =x \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n\left(\frac{k}{n}\right)^2 \\ & =x \int_0^1 x^2 d x=x\left[\frac{x^3}{3}\right]_0^1 \\ & =\frac{x}{3}\end{aligned}\)
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