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QJEEM21Ma9663: JEE Main 2021 Mathematics - Class 12 JEEM21Ma9663 Solution: The value of , where r is a non-zero real number and [r]…

The value of lim⁡n→∞[r]+[2r]+...+[nr]n2\mathop {\lim }\limits_{n \to \infty } {{[r] + [2r] + ... + [nr]} \over {{n^2}}} , where r is a non-zero real number and [r] denotes the greatest integer less than or equal to r, is equal to :

Multiple Choice Options

  • A r
  • B r2{r \over 2}
  • C 0
  • D 2r

Step-by-Step Verified Solution

We know,

(x −- 1) ≤\le [x] < x

∴\therefore (r −- 1) ≤\le [r] < r

(2r −- 1) ≤\le [2r] < 2r

.

.

.

(nr −- 1) ≤\le [nr] < nr

Adding

{{n(n + 1)} \over 2}r - n \le [r] + [2r] + .......[nr] &lt; {{n(n + 1)} \over 2}r

n(n+1)2r−nn2≤[r]+[2r]+....+[nr]n2≤n(n+1)2rn2{{{{n\left( {n + 1} \right)} \over 2}r - n} \over {{n^2}}} \le {{\left[ r \right] + \left[ {2r} \right] + .... + \left[ {nr} \right]} \over {{n^2}}} \le {{{{n\left( {n + 1} \right)} \over 2}r} \over {{n^2}}}

\mathop {\lim }\limits_{n \to \infty } \left( {{{{{n(n + 1)} \over 2}r - n} \over {{n^2}}}} \right) \le L &lt; \mathop {\lim }\limits_{n \to \infty } {{n(n + 1)} \over 2}r

\Rightarrow {r \over 2} \le L &lt; {r \over 2}

⇒L=r2\Rightarrow L = {r \over 2}

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