Search any question & find its solution
Question:
Answered & Verified by Expert
The mean of the squares of first $n$ natural numbers is
Options:
Solution:
2098 Upvotes
Verified Answer
The correct answer is:
$\frac{2 n^2+3 n+1}{6}$
First $n$ natural numbers
$$
1,2,3,4, \ldots, n
$$
Their squares $1^2, 2^2, 3^2, \ldots, n^2$
$$
\text { Now mean }=\frac{\text { Sum of observations }}{\text { Total number of observations }}
$$
$$
\begin{aligned}
& =\frac{1^2+2^2+3^2+\ldots+n^2}{n} \\
& =\frac{\frac{n}{6}(n+1)(2 n+1)}{n}\left[\because \Sigma n^2=\frac{n}{6}(n+1)(2 n+1)\right] \\
& =\frac{(n+1)(2 n+1)}{6}=\frac{2 n^2+3 n+1}{6}
\end{aligned}
$$
$$
1,2,3,4, \ldots, n
$$
Their squares $1^2, 2^2, 3^2, \ldots, n^2$
$$
\text { Now mean }=\frac{\text { Sum of observations }}{\text { Total number of observations }}
$$
$$
\begin{aligned}
& =\frac{1^2+2^2+3^2+\ldots+n^2}{n} \\
& =\frac{\frac{n}{6}(n+1)(2 n+1)}{n}\left[\because \Sigma n^2=\frac{n}{6}(n+1)(2 n+1)\right] \\
& =\frac{(n+1)(2 n+1)}{6}=\frac{2 n^2+3 n+1}{6}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.