Summation proof

• June 25th 2013, 11:57 PM
liedora
Summation proof
I have conjectured that

$\sum^{n}_{i=m}i = \sum^{n}_{i=m}( n+m-i).$

How would one go about constructing a proof for this?

Cheers.
• June 26th 2013, 12:54 AM
Shakarri
Re: Summation proof
Split the right hand side into

$\sum^{n}_{i=m}(n+m)- \sum^{n}_{i=m}(i)$

And note that $\sum^{n}_{i=m}(i)= \sum^{n}_{i=1}(i)-\sum^{m}_{i=1}(i)$
You should know how to sum integers from 1 to n
• June 26th 2013, 03:36 AM
liedora
Re: Summation proof
Got it out after I realised you made a typo :D

\displaystyle\begin{align*}\sum^{n}_{i=m}{(n+m-i)} &= \sum^{n}_{i=m}{n} + \sum^{n}_{i=m}{m} - \sum^{n}_{i=m}i \\&= \sum^{n}_{i=1}n - \sum^{m-1}_{i=1}n + \sum^{n}_{i=1}m - \sum^{m-1}_{i=1}m - \sum^{n}_{i=m}i \\&= n^2 - (m-1)n + nm - (m-1)m -\sum^{n}_{i=m}{i} \\&= n^2 + n - m^2 + m -\left[ \frac{n(n+1)}{2} - \frac{(m-1)m}{2} \right] \\& = 2 \left( \frac{n(n+1)}{2} - \frac{(m-1)m}{2} \right) - \left( \frac{n(n+1)}{2} - \frac{(m-1)m}{2}\right) \\&= \frac{n(n+1)}{2} - \frac{m(m+1)}{2} \\&= \sum^{n}_{i=1}{i} - \sum^{m-1}_{i=1}{i} \\&= \sum^{n}_{i=m}{i} \end{align*}