The sum, of the coefficients of the first 50 terms in the binomial expansion of 1 - x 100 , is equal to

We know that ${\left(1-x\right)}^{100}={}^{100}C_0-{}^{100}C_1·x+{}^{100}C_2·x^2-.......-{}^{100}C_{49}·x^{49}+.......+{}^{100}C_{100}·x^{100}$

Now let, the sum of first fifty coefficients be $S$.

$\Rightarrow S={}^{100}C_0-{}^{100}C_1+{}^{100}C_2-.......-{}^{100}C_{49}$

Let us substitute $x=1$ in the expansion.

$\Rightarrow {\left(1-1\right)}^{100}={}^{100}C_0-{}^{100}C_1·1+{}^{100}C_2·1^2-.......-{}^{100}C_{49}·1^{49}+.......+{}^{100}C_{100}·1^{100}$

$\Rightarrow 0={}^{100}C_0-{}^{100}C_1+{}^{100}C_2-.......-{}^{100}C_{49}+{}^{100}C_{50}-{}^{100}C_{51}+{}^{100}C_{52}.......+{}^{100}C_{100}$

There are total of $101$ terms and the middle term is $T_{51}={}^{100}C_{50}$.

As ${}^{n}C_r={}^{n}C_{n-r}$

$\Rightarrow 0={}^{100}C_0-{}^{100}C_1+{}^{100}C_2-.......-{}^{100}C_{49}+{}^{100}C_{50}-{}^{100}C_{49}+{}^{100}C_{48}.......+{}^{100}C_0$

$\Rightarrow 0={}^{100}C_{50}+{}^{100}C_0-{}^{100}C_1+{}^{100}C_2-.......-{}^{100}C_{49}-{}^{100}C_{49}+{}^{100}C_{48}.......+{}^{100}C_0$

$\Rightarrow 0={}^{100}C_{50}+2\left({}^{100}C_0-{}^{100}C_1+{}^{100}C_2-.......-{}^{100}C_{49}\right)$

$\Rightarrow -\frac{{}^{100}C_{50}}{2}=\left({}^{100}C_0-{}^{100}C_1+{}^{100}C_2-.......-{}^{100}C_{49}\right)$

$\Rightarrow {}^{100}C_0-{}^{100}C_1+{}^{100}C_2-.......-{}^{100}C_{49}=-\frac{{}^{100}C_{50}}{2}$

Therefore, the required answer is $-\frac{{}^{100}C_{50}}{2}=-\frac{1}{2}\times \frac{100}{50}{}^{99}C_{49}=-{}^{99}C_{49}$.