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MHT CET · Maths · Statistics

Variance of first n natural numbers is \(\qquad\) .

  1. A \(\mathrm{n}^2-\frac{1}{12}\)
  2. B \(\frac{(\mathrm{n}-1)^2}{12}\)
  3. C \(\frac{\mathrm{n}^2}{12}-1\)
  4. D \(\frac{\mathrm{n}^2-1}{12}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\mathrm{n}^2-1}{12}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \sigma^2=\frac{1}{n}\left(1^2+2^2+3^2+\ldots+n^2\right) \\ &-\left(\frac{1+2+3+\ldots+n}{n}\right)^2\end{aligned}\)
\(\begin{aligned} & =\frac{1}{n}\left(\frac{n(n+1)(2 n+1)}{6}\right)-\left(\frac{n(n+1)}{2 n}\right)^2 \\ & =\frac{(n+1)(2 n+1)}{6}-\left(\frac{n+1}{2}\right)^2 \\ & =\frac{n+1}{2}\left(\frac{2 n+1}{3}-\frac{n+1}{2}\right) \\ & =\frac{n+1}{2}\left(\frac{n-1}{6}\right) \\ & =\frac{n^2-1}{12}\end{aligned}\)
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