I'm studying integration and trying to figure out how to use the sum of squares formula to solve more complicated sums.
For example: knowing that $\sum_{i=1}^n i^2 = \frac{n (n+1) (2 n+1)}{6}$
how can we simplify
$\sum_{i=1}^n (i/n)^2$
I'm studying integration and trying to figure out how to use the sum of squares formula to solve more complicated sums.
For example: knowing that $\sum_{i=1}^n i^2 = \frac{n (n+1) (2 n+1)}{6}$
how can we simplify
$\sum_{i=1}^n (i/n)^2$
The key here is what anon wrote in his comment. It's helpful to expand out the terms of the sum: $ \begin{eqnarray} \sum_{i=1}^n (i/n)^2 & = & \sum_{i=1}^n \frac{i^2}{n^2} \\ &=& \frac{1^2}{n^2} + \frac{2^2}{n^2} + \dots + \frac{n^2}{n^2} \\ &=& \Big( \frac{{1^2} + {2^2} + \dots + {n^2}} {n^2} \Big) \\ & = & \frac{1}{n^2} \sum_{i=1}^n i^2 \\ & = & \frac{1}{n^2} \frac{n (n+1) (2 n+1)}{6} \end{eqnarray} $