Evaluate

\sum_{n = 1}^{2007} \frac{5^{2008}}{25^n + 5^{2008}}.

Solution:

Let

S = \sum_{n = 1}^{2007} \frac{5^{2008}}{25^n + 5^{2008}} = \sum_{n = 1}^{2007} \frac{5^{2008 - n}}{5^n + 5^{2008 - n}}.

Using substitution m = 2008 - n,  we have

S = \sum_{m = 1}^{2007} \frac{5^m}{5^{2008 - m} + 5^m}.

Renaming the dummy variable back to n,   we have

S = \sum_{n = 1}^{2007} \frac{5^n}{5^n + 5^{2008 - n}}.

Hence

2S = \sum_{n = 1}^{2007} \frac{5^n}{5^n + 5^{2008 - n}} + \sum_{n = 1}^{2007} \frac{5^{2008 - n}}{5^n + 5^{2008 - n}} = \sum_{n = 1}^{2007}{1} = 2007.

It follows that S = \frac{2007}{2}.


Reference

  1. Zengxiang Tong, S126. A simple Sum, Math Magazine, MAA, Nov. 2008, p.33