Question

If [x] denotes the greatest integer $\le x,then\underset{x\to \infty }{lim}\frac{1}{n^3}\left\{\right[1^{2}x]+[2^{2}x]+[3^{2}x]+ldots\dots +[n^{2}x\left]\right\}$ equals

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$\underset{x\to \infty }{lim}\frac{1}{n^3}\left\{\right[1^{2}x]+[2^{2}x]+[3^{2}x]+ldots\dots +[n^{2}x\left]\right\}=\underset{x\to \infty }{lim}\left\{\frac{{\Sigma }_{r=1}^n\left[r^{2}x\right]}{n^3}\right\}=\underset{x\to \infty }{lim}\left(\frac{{\Sigma }_{r=1}^n{r}^{2}x-\left\{r^{2}x\right\}}{n^3}\right)=\underset{x\to \infty }{lim}(\frac{x.\frac{n(n+1)(2n+1)}{6}}{n^3}-{\Sigma }_{r=1}^n\frac{\left\{r^{2}x\right\}}{n^3})=x.\frac{\left(1\right)\left(1\right)\left(2\right)}{6}-0=\frac{x}{3}$