An algebra problem by neelesh vij

Intelligently set question!! But, only a single observation would ease the question. Let sum of the given expression be S. And as in summation we can replace the summation variable by [lower limit + upper limit - summation variable] i.e (x by 2009+0-a), the sum remains the same (S). Now adding both of the expressions we would see that numerator and denominator cancels and if we are left with summation of 1 under the given limits and calculating that we would get 2S=2010 i.e S=1005

By putting (2009-x) in f(x) we get,

f(2009-x) = $\frac{2009-x}{x}$

or, f(2009-x) = 1 - $\frac{x}{2009}$

Therefore,

f(2009-x) + f(x)=1 ........ (1)

Now,

putting (2009-x) in g(x) function we get,

g(2009-x)

= \frac{\(\f(2009-x))^{4}}{\(\f(2009-x))^{4} + \(\f(x))^{4}} = \frac{\(\1 - f(x))^{4}}{\(\f(x))^{4} + \(\1 - f(x))^{4}} [Since , f(2009-x) = 1 - f(x) from 1]

We observre, g(x) + g(2009-x) = 1
So, $\displaystyle \sum_{x=0}^{2009} g(x)$

= [g(0) + g(2009)] + [g(1) + g(2008)] + ............. + [g(1004) + g(1005)]

= 1*1005

=1005