1no i guess eragon is correct....
If f(x)=\frac{a^x}{a^x +\sqrt{a}} [a>0]
then evaluate
\sum_{r=1}^{2n-1}{2f(\frac{r}{2n})}
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4 Answers
1if we take any i\epsilon (1,2n-1)
f\left(\frac{i}{2n} \right)+f\left(\frac{2n-i }{2n}\right)=1
hence
f\left(\frac{1}{2n} \right)+f\left(\frac{2n-1 }{2n}\right)=1
f\left(\frac{2}{2n} \right)+f\left(\frac{2n-2 }{2n}\right)=1
so on
therefore in sumation \sum_{r=1}^{2n-1}{f\left(\frac{r}{2n} \right)}
sum of first and last term is1, 2nd and second last term is 1 and so on
this leaves us with middle term which is {f\left(\frac{n}{2n} \right)}
hence teh given summation is 2\sum_{r=1}^{2n-1}{}{f\left(\frac{r}{2n} \right)}=2(n-1)+2f(\frac{1}{2})=2(n-1)+2.\frac{1}{2}=2n-1
