Stubborn Odd Binomial Coefficient

$f(x) = (1+x)^{25} = C_0 + C_1 x + C_2 x^2 + ... + C_{25} x^{25}$ and $g(x) = (1-x)^{25} = C_0 - C_1 x + C_2 x^2 - ... - C_{25} x^{25}$

$h(x) = f(x)-g(x) = (1+x)^{25} - (1-x)^{25} = 2( C_1 x + C_3 x^3 - ... - C_{25} x^{25} )$

then differentiating w.r.t. x,

$h'(x) = f'(x)-g'(x) = 25(1+x)^{24} + 25(1-x)^{24} = 2( C_1 + 3 C_3 x^2 - ... - 25 C_{25} x^{24} )$

Evaluate h'(x) at x=i where $i=\sqrt {-1} ...$

h'(i) = f'(i) - g'(i) = $25(1+i)^{24} + 25(1-i)^{24}$ = $2( C_1 + 3 C_3 i^2 - ... - 25 C_{25} i^{24} )$ $= 2( C_1 - 3 C_3 + ... - 25 C_{25} )$

Since:

$25(1+i)^{24}$ $= 25 [ \sqrt 2 ( \cos{(\pi/4)} + i \sin{(\pi/4))} ]^{24}$ $= 25 (2^{12}) [ \cos{(6\pi)} + i \sin{(6\pi)} ] = 102400$

and

$25(1-i)^{24}$ $= 25 [ \sqrt 2 ( \cos{(-\pi/4)} + i \sin{(-\pi/4))} ]^{24}$ $= 25 (2^{12}) [ \cos{(-6\pi)} + i \sin{(-6\pi)} ] = 102400$

So

h'(i) = 102400 + 102400

$C_1 - 3 C_3 + 5 C_5 -... - 25 C_25 = h'(i)/2 = \boxed {102,400}$