The average of the squares of the numbers 0,1,2, ,n is

Correct Answer is : 16n(2n+1)1/6 n(2n+1)61n(2n+1)

Correct Answer is : 16n(2n+1)1/6 n(2n+1)61n(2n+1)

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UPSC NDA Math Model Paper 6

Question : 26 of 120

Marks: +1, -0

0,1,2,\ldots ,n

Solution:

Here, we have to find the average of the squares of the numbers 0, 1, 2, ......, n

First let's find

\sum\limits_{i=1}^{n+1} x_{i}^{2}=0^{2}+1^{2}+2^{2}+\ldots+n^{2}

As we know that,

1^{2}+2^{2}+3^{2}+\dots+n^{2}

=\sum n^{2}=\frac{n(n+1)(2n+1)}{6}

\Rightarrow \sum\limits_{i=1}^{n+1} x_{i}^{2}=\frac{n \times (n+1) \times (2n+1)}{6}

As we know that, the average of

n

observations

x_{1}, x_{2}, \ldots, x_{n}

is given by:

\frac{\sum\limits_{i=1}^{n} x_{i}}{n}

\Rightarrow \frac{\sum\limits_{i=1}^{n+1} x_{i}^{2}}{n+1} = \frac{ \left[ \frac{n \times (n+1) \times (2n+1)}{6} \right] }{n+1}

=\frac{n(2n+1)}{6}

Hence, option B is the correct answer.

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