Welcome 2016! Part 31

$\begin{aligned} \text{We note that:} \quad (1+x)^n & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k \\ \text{When }x = i \Rightarrow (1+i)^n & = 1 + \begin{pmatrix} n \\ 1 \end{pmatrix} i + \begin{pmatrix} n \\ 2 \end{pmatrix} i^2 + \begin{pmatrix} n \\ 3 \end{pmatrix} i^3 + \begin{pmatrix} n \\ 4 \end{pmatrix} i^4 + \begin{pmatrix} n \\ 5 \end{pmatrix} i^5 +...+ \begin{pmatrix} n \\ n \end{pmatrix} i^n \\ & = 1 + \begin{pmatrix} n \\ 1 \end{pmatrix} i - \begin{pmatrix} n \\ 2 \end{pmatrix} - \begin{pmatrix} n \\ 3 \end{pmatrix} i + \begin{pmatrix} n \\ 4 \end{pmatrix} + \begin{pmatrix} n \\ 5 \end{pmatrix} i +...+ \begin{pmatrix} n \\ n \end{pmatrix} i^n \\ \text{Therefore,} \quad \sum_{n=0}^{49} \begin{pmatrix} 99 \\ 2n \end{pmatrix} & = \Re \left \{ (1+i)^{99} \right \} \quad \quad \color{#3D99F6}{\text{As }(1+i)^2 = 2i} \\ & = \Re \left \{ (2i)^{49}(1+i) \right \} \\ & = \Re \left \{ 2^{49}i(1+i) \right \} \\ & = \Re \left \{ 2^{49}(i-1) \right \} \\ & = - 2^{49} \end{aligned}$

$\Rightarrow a + b = -2+49 = \boxed{47}$

$(1+i)^{99} = \left[ {99 \choose 0} - {99 \choose 2} + {99 \choose 4} - ... \right ] + i \left [ {99 \choose 1} - {99 \choose 3} + ... \right ]$

Thus we have $(1+i)^{99} = -2^{49} + i2^{49}$

Comparing real components, we have that the desired expression is $(-2)^{49}$ .