the function f(x) is such that, f(x) + f(1-x) = k where k is a constant, then , prove that SIGMA(r=1,m)f(r/m+1) = mk/2
$\displaystyle \displaystyle S=\sum^{m}_{r=1}f\left(\frac{r}{m+1}\right)=f\left (\frac{1}{m+1}\right)+f\left(\frac{2}{m+1}\right)+ \dots+f\left(\frac{m}{m+1}\right).$
$\displaystyle \displaystyle x\equiv\frac{r}{m+1}\quad\mbox{ so }\quad 1-x=1-\frac{r}{m+1}=\frac{m+1-r}{m+1
}.$
$\displaystyle \displaystyle f(x)+f(1-x)=k\quad\mbox{ so }\quad f(x)=k-f(1-x),$
so alternatively,
$\displaystyle \displaystyle S=\sum^{m}_{r=1}\left\{k-f\left(\frac{m+1-r}{m+1}\right)\right\}$
$\displaystyle \displaystyle=\left\{k-\left(\frac{m}{m+1}\right)\right\}+\left\{k-f\left(\frac{m-1}{m+1}\right)\right\}+\dots+\left\{k-f\left(\frac{1}{m+1}\right)\right\}.$
Now add the two expressions for $\displaystyle S$ together and divide by two.