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October 23rd, 2017, 10:14 AM   #1
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Post Help!

I've been trying to solve this for ages now. Any ideas what to do?
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October 23rd, 2017, 10:32 AM   #2
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Here's (a), the rest you'll have to work out on your own if you want to learn anything.

$\begin{align*}

\displaystyle

&\sum_{k=1}^n = \\

&\dfrac 1 2 \left( \sum_{k=1}^n + \sum_{k=1}^n \right) = \\

&\dfrac 1 2 \left( \sum_{k=1}^n + \sum_{k=n}^1 \right) = \\

&\dfrac 1 2 \sum_{k=1}^n~(k + n+1-k) = \\

&\dfrac 1 2 \sum_{k=1}^n ~(n+1) = \\

&\dfrac 1 2 n(n+1)

\end{align*}$
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