Question

If [x] denotes the greatest integer $\le$ x, then evaluate ${lim}_{n\to \infty }\frac{1}{n^3}\left\{\right[1^{2}x]+[2^{2}x]+[3^{2}x+.....+\left[n^{2}x\right]\}$

• A. $\frac{x}{2}$
  
• B. $\frac{x}{3}$
  
• C. $\frac{x}{4}$
  
• D. x

Answer 1

The correct option is B

$\frac{x}{3}$

${lim}_{n\to \infty }\frac{1}{n^3}\left\{\right[1^{2}x]+[2^{2}x]+[3^{2}x+.....+\left[n^{2}x\right]\}={lim}_{n\to \infty }\left\{\frac{{\sum }_{r=1}^n\left[r^{2}x\right]}{n^3}\right\}={lim}_{n\to \infty }\left\{\frac{{\sum }_{r=1}^n{r}^{2}x-\left\{r^{2}x\right\}}{n^3}\right\},where\left\{\right\}denotesfractionalpartfunction={lim}_{n\to \infty }\{\frac{x\frac{n(n+1)(2n+1)}{6}}{n^3}-{\sum }_{r=1}^n\frac{\left(r^{2}x\right)}{n^3}\}=x\frac{\left(1\right)\left(1\right)\left(2\right)}{6}-0=\frac{x}{3}$