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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

$\sum_{i=1}^n (i/n)^2$

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    $\displaystyle\sum_i c\,a_i=c \sum_i a_i$ whenever $c$ is a constant (meaning not dependent on the index $i$) So you can pull $n^2$ out of the sum and use the formula you have.2012-08-20

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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} $