jmc

By the Binomial Theorem, $$\begin{array}{rl}(1+i{)}^{99}& =\left(\genfrac{}{}{0ex}{}{99}{0}\right)+\left(\genfrac{}{}{0ex}{}{99}{1}\right)i+\left(\genfrac{}{}{0ex}{}{99}{2}\right)i^2+\left(\genfrac{}{}{0ex}{}{99}{3}\right)i^3+\cdots +\left(\genfrac{}{}{0ex}{}{99}{98}\right)i^{98}+\left(\genfrac{}{}{0ex}{}{99}{99}\right)i^{99}\\ & =\left(\genfrac{}{}{0ex}{}{99}{0}\right)+\left(\genfrac{}{}{0ex}{}{99}{1}\right)i-\left(\genfrac{}{}{0ex}{}{99}{2}\right)-\left(\genfrac{}{}{0ex}{}{99}{3}\right)i+\cdots -\left(\genfrac{}{}{0ex}{}{99}{98}\right)-\left(\genfrac{}{}{0ex}{}{99}{99}\right)i.\end{array}$$Thus, the sum we seek is the real part of $(1+i{)}^{99}.$

Note that $(1+i{)}^2=1+2i+i^2=2i,$ so $$\begin{array}{rl}(1+i{)}^{99}& =(1+i{)}^{98}\cdot (1+i)\\ & =(2i{)}^{49}(1+i)\\ & =2^{49}\cdot i^{49}\cdot (1+i)\\ & =2^{49}\cdot i\cdot (1+i)\\ & =2^{49}(-1+i)\\ & =-2^{49}+2^{49}i.\end{array}$$Hence, the given sum is $\boxed{-2^{49}}.$