jmc

By the Binomial Theorem, $(1+i{)}^{99}={\sum }_{n=0}^{99}(\genfrac{}{}{0ex}{}{99}{j})i^n=$ $\left(\genfrac{}{}{0ex}{}{99}{0}\right)i^0+\left(\genfrac{}{}{0ex}{}{99}{1}\right)i^1+\left(\genfrac{}{}{0ex}{}{99}{2}\right)i^2+\left(\genfrac{}{}{0ex}{}{99}{3}\right)i^3+\left(\genfrac{}{}{0ex}{}{99}{4}\right)i^4+\cdots +\left(\genfrac{}{}{0ex}{}{99}{98}\right)i^{98}$. Using the fact that $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, and $i^{n+4}=i^n$, the sum becomes: $(1+i{)}^{99}=(\genfrac{}{}{0ex}{}{99}{0})+(\genfrac{}{}{0ex}{}{99}{1})i-(\genfrac{}{}{0ex}{}{99}{2})-(\genfrac{}{}{0ex}{}{99}{3})i+(\genfrac{}{}{0ex}{}{99}{4})+\cdots -(\genfrac{}{}{0ex}{}{99}{98})$. So, $Re[(1+i{)}^{99}]=(\genfrac{}{}{0ex}{}{99}{0})-(\genfrac{}{}{0ex}{}{99}{2})+(\genfrac{}{}{0ex}{}{99}{4})-\cdots -(\genfrac{}{}{0ex}{}{99}{98})=S$. Using De Moivre's Theorem, $(1+i{)}^{99}=[\sqrt{2}cis(45^{\circ }){]}^{99}=\sqrt{2^{99}}\cdot cis(99\cdot 45^{\circ })=2^{49}\sqrt{2}\cdot cis(135^{\circ })=-2^{49}+2^{49}i$. And finally, $S=Re[-2^{49}+2^{49}i]=\boxed{-2^{49}}$.