Re: HSC 2013 4U Marathon
Not necessarily starting from 1, 2, 3
Generally from r, r+1 etc. But its not worth answering lol its very simple, I just thought it was a cool result.
I was able to arrive at:
\sum_{k=2}^{n} 1 - \frac{2}{k+1} = \frac{n}{2}(n-1) - 2\sum_{k=2}^{n} \frac{1}{k+1}
And...