October 23rd, 2017, 10:14 AM  #1 
Newbie Joined: Oct 2017 From: Sylhet, Bangladesh Posts: 1 Thanks: 0  Help!
I've been trying to solve this for ages now. Any ideas what to do?

October 23rd, 2017, 10:32 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 
Here's (a), the rest you'll have to work out on your own if you want to learn anything. $\begin{align*} \displaystyle &\sum_{k=1}^n = \\ &\dfrac 1 2 \left( \sum_{k=1}^n + \sum_{k=1}^n \right) = \\ &\dfrac 1 2 \left( \sum_{k=1}^n + \sum_{k=n}^1 \right) = \\ &\dfrac 1 2 \sum_{k=1}^n~(k + n+1k) = \\ &\dfrac 1 2 \sum_{k=1}^n ~(n+1) = \\ &\dfrac 1 2 n(n+1) \end{align*}$ 