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

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Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { } \lim _{n ightarrow \infty} \frac{1}{n^3}\left\{\left[1^2 xight]+\left[2^2 xight]+\left[3^2 xight]+\ldots+\left[n^2 xight]ight\} \ & =\lim _{n ightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n\left[r^2 xight]=\lim _{n ightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n\left(r^2 x-\left\{r^2 xight\}ight) \ & =\lim _{n ightarrow \infty} \frac{1}{n^3}\left(x \sum_{r=1}^n r^2-\sum_{r=1}^n\left\{r^2 night\}ight) \ & =\lim _{n ightarrow \infty} \frac{x 	imes n(n+1)(2 n+1)}{n^3 	imes 6}-\sum_{r=1}^n \frac{\left\{r^2 night\}}{n^3} \ & =\lim _{n ightarrow \infty}\left(\frac{x}{6} 	imes \frac{n}{n} 	imes \frac{n+1}{n} 	imes \frac{(2 n+1)}{n}ight)-\lim _{n ightarrow \infty} \sum_{r=1}^n \frac{\left\{r^2 night\}}{n^3} \ & =\frac{x}{6} 	imes 1 	imes 2-0=\frac{x}{3} \end{aligned} $

If $[x]$ denotes the greatest integer $\leq x$, then $\lim _{n \rightarrow-\infty}…

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