MHT CET · Maths · Statistics
Variance of first n natural numbers is \(\qquad\) .
- A \(\mathrm{n}^2-\frac{1}{12}\)
- B \(\frac{(\mathrm{n}-1)^2}{12}\)
- C \(\frac{\mathrm{n}^2}{12}-1\)
- D \(\frac{\mathrm{n}^2-1}{12}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{n}^2-1}{12}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sigma^2=\frac{1}{n}\left(1^2+2^2+3^2+\ldots+n^2\right) \\ &-\left(\frac{1+2+3+\ldots+n}{n}\right)^2\end{aligned}\)
\(\begin{aligned} & =\frac{1}{n}\left(\frac{n(n+1)(2 n+1)}{6}\right)-\left(\frac{n(n+1)}{2 n}\right)^2 \\ & =\frac{(n+1)(2 n+1)}{6}-\left(\frac{n+1}{2}\right)^2 \\ & =\frac{n+1}{2}\left(\frac{2 n+1}{3}-\frac{n+1}{2}\right) \\ & =\frac{n+1}{2}\left(\frac{n-1}{6}\right) \\ & =\frac{n^2-1}{12}\end{aligned}\)
\(\begin{aligned} & =\frac{1}{n}\left(\frac{n(n+1)(2 n+1)}{6}\right)-\left(\frac{n(n+1)}{2 n}\right)^2 \\ & =\frac{(n+1)(2 n+1)}{6}-\left(\frac{n+1}{2}\right)^2 \\ & =\frac{n+1}{2}\left(\frac{2 n+1}{3}-\frac{n+1}{2}\right) \\ & =\frac{n+1}{2}\left(\frac{n-1}{6}\right) \\ & =\frac{n^2-1}{12}\end{aligned}\)
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