help plz

sumfor(n, 0, 49, (-1)^n * (99  nCr  (2*n))=-562,949,953,421,312 - A very strange result !!.

Find the value of the sum \(\dbinom{99}{0} - \dbinom{99}{2} + \dbinom{99}{4}- \dbinom{99}{6} + \dots - \dbinom{99}{98}\)

\((1+i)^{99}=\dbinom{99}{0}+i\dbinom{99}{1}-\dbinom{99}{2}-i\dbinom{99}{3}+\dbinom{99}{4}+i\dbinom{99}{5}- \ldots -\dbinom{99}{98}-i\dbinom{99}{99} \qquad (1)\)

Note here that \(\binom{99}{0} - \binom{99}{2} + \binom{99}{4}- \binom{99}{6} + \dots - \binom{99}{98}\) is the real part of (1).

\(\begin{array}{|rcll|} \hline 1+i &=& \sqrt{1^2+1^2} \Bigg(\cos\Big( \arctan(\frac{1}{1} )\Big) +i\sin\Big( \arctan(\frac{1}{1} )\Big) \Bigg) \\ 1+i &=& \sqrt{2} \Big(\cos(45^\circ) +i\sin(45^\circ) \Big) \\ \hline (1+i)^{99} &=& \Bigg( \sqrt{2} \Big(\cos(45^\circ) +i\sin(45^\circ) \Big) \Bigg)^{99} \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(\cos(99*45^\circ) +i\sin(99*45^\circ) \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(\cos(4455^\circ) +i\sin(4455^\circ) \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(\cos(135^\circ) +i\sin(135^\circ) \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(-\dfrac{\sqrt{2}}{2} +i\dfrac{\sqrt{2}}{2} \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(-2^{\frac{1}{2}-1} +i\dfrac{\sqrt{2}}{2} \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(-2^{-\frac{1}{2}} +i\dfrac{\sqrt{2}}{2} \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}}*(-2^{-\frac{1}{2}}) +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& -2^{\frac{99}{2}}2^{-\frac{1}{2}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& -2^{\frac{99}{2}-\frac{1}{2}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& -2^{\frac{98}{2}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& \underbrace{\color{red}-2^{49}}_{\text{the real part}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ \hline \end{array} \)

\(\dbinom{99}{0} - \dbinom{99}{2} + \dbinom{99}{4}- \dbinom{99}{6} + \dots - \dbinom{99}{98} = \mathbf{-2^{49}}\)