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NDA 2013
Exam
NDA
Year
2013
Subject
Mathematics
Chapter
Sequences and Series

NDA 2013 · Mathematics · Sequences and Series

The arithmetic mean of the squares of the first n natural numbers is

The arithmetic mean of the squares of the first nn natural numbers is

Options

  1. An(n+1)(2n+1)6\frac{n(n+1)(2 n+1)}{6}
  2. Bn(n+1)(2n+1)2\frac{n(n+1)(2 n+1)}{2}
  3. C(n+1)(2n+1)6\frac{(n+1)(2 n+1)}{6}
  4. D(n+1)(2n+1)3\frac{(n+1)(2 n+1)}{3}

Correct answer

C. (n+1)(2n+1)6\frac{(n+1)(2 n+1)}{6}

Step-by-step solution

A.M.A.M.= sum of square of 1st n natural numbers n=n(n+1)(2n+1)6n=(n+1)(2n+1)6\begin{aligned} & =\frac{\text{ sum of square of 1st } n \text{ natural numbers }}{n} \\ & =\frac{n(n+1)(2 n+1)}{6 n}=\frac{(n+1)(2 n+1)}{6} \end{aligned}$$
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