If . denote the greatest integer function then Ltn→∞x+2x+3x+......+nxn2=

Ltn→∞x+2x+3x+.....+nxn2=Ltn→∞x−x+2x−x+3x−3x+....+nx−nxn2=Ltn→∞x1+2+3+...+nn2−x+2x+....n2=Ltn→∞xnn+12n2−0=x2

If . denote the greatest integer function then Ltn→∞x+2x+3x+......+nxn2=

Ltn→∞x+2x+3x+.....+nxn2=Ltn→∞x−x+2x−x+3x−3x+....+nx−nxn2=Ltn→∞x1+2+3+...+nn2−x+2x+....n2=Ltn→∞xnn+12n2−0=x2

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$If [.] denote the greatest integer function then \underset{n \to \infty}{Lt} \frac{[x] + [2 x] + [3 x] + . . . . . . + [nx]}{n^{2}} =$

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$\frac{x}{2}$

$\frac{x}{3}$

$\frac{x}{6}$

answer is A.

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Detailed Solution

$\begin{array}{l} \underset{n \to \infty}{L t} \frac{[x] + [2 x] + [3 x] + ..... + [n x]}{n^{2}} \\ = \underset{n \to \infty}{L t} \frac{x - \{x\} + 2 x - \{x\} + 3 x - \{3 x\} + .... + n x - \{n x\}}{n^{2}} \\ = \underset{n \to \infty}{L t} \frac{x (1 + 2 + 3 + ... + n)}{n^{2}} - \frac{\{x\} + \{2 x\} + ....}{n^{2}} \\ = \underset{n \to \infty}{L t} \frac{x \frac{n (n + 1)}{2}}{n^{2}} - 0 = \frac{x}{2} \end{array}$