How do you use sigma notation to write the sum for [1-(1/6)^2]+[1-(2/6)^2]+...+[1-(6/6)^2]?

Apr 6, 2017

${\sum}_{n = 1}^{6} \left(1 - {\left(\frac{n}{6}\right)}^{2}\right)$
$\left[1 - {\left(\frac{\textcolor{red}{1}}{6}\right)}^{2}\right] + \left[1 - {\left(\frac{\textcolor{red}{2}}{6}\right)}^{2}\right] + \ldots + \left[1 - {\left(\frac{\textcolor{red}{6}}{6}\right)}^{2}\right]$.
It starts from one and is incremented by one until it reaches 6. Everything else stays the same. We just need to substitute the number that is changing by a variable: ${\sum}_{n = 1}^{6} \left(1 - {\left(\frac{n}{6}\right)}^{2}\right)$.