Very alternating binomial coefficients

Expand $(1+i)^{99} = \left( \binom{99}{0}-\binom{99}{2} + \binom{99}{4} - \cdots \right) + i \left( \binom{99}{1}-\binom{99}{3}+\cdots\right)$ by the binomial theorem , where we have grouped real and imaginary terms together.

So the sum we want is just the real part of $(1+i)^{99}$ . This power equals $(2i)^{49}(1+i) =2^{49}i(1+i) = 2^{49}(-1+i).$ The real part is $-2^{49} = (-2)^{49}$ , so the answer is $-2+49 = \fbox{47}$ .

((1+i)^99+(1-i)^99)/2= 99C0-99C2+99C4...

=2^(99/2)(e^(i99π/4)+e^(-i99π/4))/2

=2^(97/2)(2cos(99π/4))

=-2^49

=(-2)^49

a+b=49-2=47

I just used Pascal's triangle and saw a pattern for powers of 2 in the third, fifth, seventh and ninth rows. It was (-2)^(k-1)/2