Question

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

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

Answer 1

Answer: (B)

$\frac{x}{3}$

For any integer, $k$ we have

$k^{2}x-1<\left[k^{2}x\right]\le k^{2}x+1$

$\Rightarrow {\sum }_{k=1}^n(k^{2}x-1)<{\sum }_{k=1}^n\left[k^{2}x\right]\le {\sum }_{k=1}^n(k^{2}x+1)$

$\Rightarrow \frac{1}{n^3}{\sum }_{k=1}^n(k^{2}x-1)<\frac{1}{n^3}{\sum }_{k=1}^n\left[k^{2}x\right]\le \frac{1}{n^3}{\sum }_{k=1}^n(k^{2}x+1)$

$\Rightarrow \frac{1}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)x-\frac{n}{n^3}<\frac{1}{n^3}{\sum }_{k=1}^n\left[k^{2}x\right]\le \left(\frac{n(n+1)(2n+1)}{6}\right)x+\frac{n}{n^3}$

$\Rightarrow \left(\frac{2n^2+3n+1}{6n^2}\right)x-\frac{1}{n^2}<\frac{1}{n^3}{\sum }_{k=1}^n\left[k^{2}x\right]\le \left(\frac{2n^2+3n+1}{6n^2}\right)+\frac{1}{n^2}$

$\frac{x}{3}-0<{lim}_{n\to \infty }\frac{1}{n^3}{\sum }_{k=1}^n\left[k^{2}x\right]\le \frac{x}{3}$

$\Rightarrow {lim}_{n\to \infty }\frac{1}{n^3}{\sum }_{k=1}^n\left[k^{2}x\right]=\frac{x}{3}$