Calculate $\sum_{k=1}^{2007} \frac{5^{2008}}{25^k+5^{2008}}$

Question

$\sum_{k=1}^{2007} \frac{5^{2008}}{25^k+5^{2008}}$. I know that solution is $\frac{2007}{2}$ and that I can use a Gaussian method.

If I use Gaussian method I do not get some sum that help me. if I put $n=2008$, then $Sn=\sum_{k=1}^n \frac{5^{n+1}}{25^k+5^{n+1}}$. and $Sn=\sum_{k=1}^n \frac{5^{n}}{25^{k-n}+5^{n+1}}$. So then $2Sn=\sum_{k=1}^n \frac{5^{n+1}}{25^k+5^{n+1}}+\frac{5^{n+1}}{25^{n-k}+5^{n+1}}$. do you have something better?

Answer 1

Using $\frac{1}{1+q}+\frac{1}{1+1/q}=1$, $$\sum_{k=1}^n\frac{1}{1+5^{2k-n-1}}=\frac{1}{2}\sum_{k=1}^n\bigg(\frac{1}{1+5^{2k-n-1}}+\frac{1}{1+5^{2(n+1-k)-n-1}}\bigg)\\=\frac{1}{2}\sum_{k=1}^n\bigg(\frac{1}{1+5^{2k-n-1}}+\frac{1}{1+5^{n+1-2k}}\bigg)=\frac{1}{2}\sum_{k=1}^n 1=\frac{n}{2}.$$

Answer 2

Write $$\sum_{k=1}^{2007} \frac{5^{2008}}{25^k+5^{2008}}=\sum_{k=1}^{2007}{25^{1004}\over 25^k+25^{1004}} = \sum_{k=1}^{2007}{1\over 25^{k-1004}+1} =\sum_{j=-1003}^{1003}{1\over 2^{-j}+1}$$

Now try your pairing strategy.