If [x] denotes the greatest integer leq x,the | vedclass

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(B) We know that $[r^2 x] = r^2 x - \{r^2 x\}$,where $\{r^2 x\}$ is the fractional part of $r^2 x$. Substituting this into the limit,we get: $\lim_{n

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$\frac{x}{3}$

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(B) We know that $[r^2 x] = r^2 x - \{r^2 x\}$,where $\{r^2 x\}$ is the fractional part of $r^2 x$. Substituting this into the limit,we get: $\lim_{n \rightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n [r^2 x] = \lim_{n \rightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n (r^2 x - \{r^2 x\})$ $= \lim_{n \rightarrow \infty} \left( \frac{x}{n^3} \sum_{r=1}^n r^2 - \frac{1}{n^3} \sum_{r=1}^n \{r^2 x\} \right)$ Using the formula $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$,we have: $= \lim_{n \rightarrow \infty} \left( \frac{x \cdot n(n+1)(2n+1)}{6n^3} - \frac{1}{n^3} \sum_{r=1}^n \{r^2 x\} \right)$ Since $0 \leq \{r^2 x\} Thus,the limit is $\lim_{n \rightarrow \infty} \frac{x(2n^3 + 3n^2 + n)}{6n^3} = \frac{2x}{6} = \frac{x}{3}$.

If $[x]$ denotes the greatest integer $\leq x$,then $\lim_{n \rightarrow \infty} \frac{1}{n^3} \{[1^2 x] + [2^2 x] + [3^2 x] + \ldots + [n^2 x] \} = $

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