(a-b)^n=a^n-nba^(n-1)+n(n-1)b^2a^(n-2)-... which can also be written a^n*nC0-ba^(n-1)*nC1+b^2a^(n-2)*nC2-...,
where the mathematical symbol for combination of r distinct items out of n, nCr=C (n،r).
If a=b=1 and n=100 we have (1-1)^100=0=100C0-100C1+100C2-100C3+...-100C99+100C100.
Therefore, taking all the negative terms to the other side of the equation, 100C1+100C3+100C5+...100C99=100C0+100C2+100C4+...100C100 QED.