Question

If [x] denotes the greatest integer less than or equal to $x$ then $\underset{n\to \infty }{lim}\frac{1}{n^3}\left\{\right[1^{2}x]+[2^{2}x]+[3^{2}x]+....+[n^{2}x\left]\right\}=$

• A. $\frac{x}{2}$
• B. $\frac{x}{3}$
• C. $\frac{x}{6}$
• D. $0$

Answer 1

The correct option is B $\frac{x}{3}$

$li{m}_{x\to \infty }\frac{1}{n^3}\left\{\right[1^{2}x]+[2^{2}x]+[3^{2}x]++[n^{2}x\left]\right\}={lim}_{n\to \infty }\frac{1}{n^3}{\sum }_{k=1}^n\left[k^{2}x\right]$
As $k^{2}x-1<\left[k^{2}x\right]\Rightarrow {\sum }_{k=1}^n(k^{2}x+1)<{\sum }_{k=1}^n\left[k^{2}x\right]<{\sum }_{k=1}^n(k^{2}x+1)\Rightarrow \frac{xn(n+1)(2n+1)}{6}-n<{\sum }_{k=1}^n\left[k^{2}x\right]<\frac{xn(n+1)(2n+1)}{6}+n\Rightarrow \frac{x}{6}(1+\frac{1}{n})(2+\frac{1}{n})-\frac{1}{n^2}<\frac{1}{n^3}{\sum }_{k=1}^n\left[k^{2}x\right]<\frac{x}{6}(1+\frac{1}{n})(2+\frac{1}{n})+\frac{1}{n^2}\Rightarrow {lim}_{n\to \infty }(\frac{x}{6}(1+\frac{1}{n})(2+\frac{1}{n})-\frac{1}{n^2})<{lim}_{n\to \infty }\frac{1}{n^3}{\sum }_{k=1}^n\left[k^{2}x\right]<{lim}_{n\to \infty }(\frac{x}{6}(1+\frac{1}{n})(2+\frac{1}{n})-\frac{1}{n^2})\Rightarrow \frac{x}{3}<{lim}_{n\to \infty }\frac{1}{n^3}{\sum }_{k=1}^n\left[k^{2}x\right]<\frac{x}{3}\Rightarrow {lim}_{n\to \infty }\frac{1}{n^3}{\sum }_{k=1}^n\left[k^{2}x\right]=\frac{x}{3}$
Hence, option 'B' is correct.