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Feb 6, 2015 at 0:24 vote accept roro172
Feb 5, 2015 at 21:40 comment added Mark Counting upwards from $x=r$ to $x=n$ is equivalent to counting downwards from $j=n-r$ to $j=0$ ($j$ being the 'distance' between $x$ and $n$). You can just substitute $j = n - x$ into the limits of the sum: $\sum_{x=r}^n \equiv \sum_{j=n-r}^{n-n} \equiv \sum_{j=n-r}^0 \equiv \sum_{j=0}^{n-r}$
Feb 5, 2015 at 20:59 comment added roro172 I am still unclear on how you managed to change the upper and lower bounds of the sum?
Feb 5, 2015 at 5:29 history answered Mark CC BY-SA 3.0