Q59 / 90 · JEE Main 2012 (12 May Online) · Mathematics

A number $n$ is randomly selected from the set $\{1,2,3, \ldots ., 1000\}$. The probability that $\frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}$ is an integer is

  • A
    $0.331$
  • B
    $0.333$
  • Correct answer
    C
    $0.334$
  • D
    $0.332$

Solution

$\frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}=\frac{\frac{n(n+1)(2 n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2 n+1}{3}$ For $n=1,2,3, \ldots . ., 1000$ Value of $\frac{2 n+1}{3}=\frac{3}{3}, \frac{5}{3}, \frac{7}{3}, \ldots \ldots \ldots, \frac{2001}{3}$ respectively. Out of $\frac{3}{3}, \frac{5}{3}, \frac{7}{3}, \ldots \ldots \ldots, \frac{2001}{3}$ only first term $\left(\frac{3}{3}=1\right)$, fourth term $\left(\frac{9}{3}=3\right), 667^{\text {th }}$ term $\left(\frac{2001}{3}=667\right)$ are integers. Hence, out of 1000 values of $\frac{2 n+1}{3}$, total number of integral values of $\frac{2 n+1}{3}$ $=333+1=334$ $\therefore \quad$ Required probability $=\frac{334}{1000}=0.334$

Hence, the correct answer is option: 'C'.

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