The sum, of the coefficients of the first 50 terms in the binomial expansion of ${(1-\mathrm{x})}^{100}$ , is equal to 

Sum of the coefficients of the first 50 terms in the expansion of ${(1-\mathrm{x})}^{100}$
$={100}_{{\mathrm{C}}_0}-{100}_{{\mathrm{C}}_1}+{100}_{{\mathrm{C}}_2}-{100}_{{\mathrm{C}}_3}.......-{100}_{{\mathrm{C}}_{49}}$
We know that,
$({100}_{{\mathrm{C}}_0}-{100}_{{\mathrm{C}}_1}+{100}_{{\mathrm{C}}_2}.......-{100}_{{\mathrm{C}}_{49}})+({100}_{{\mathrm{C}}_{50}}-{100}_{{\mathrm{C}}_{51}}.......+{100}_{{\mathrm{C}}_{100}})=0$
Now $2[{}^{100}\mathrm{C}_0-{}^{100}\mathrm{C}_1+{}^{100}\mathrm{C}_2........-{}^{100}\mathrm{C}_{49}]+{}^{100}\mathrm{C}_{50}=0$ $(\because {}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}={}^{\mathrm{n}}\mathrm{C}_{\mathrm{n}-\mathrm{r}})$
$\therefore {}^{100}\mathrm{C}_0-{}^{100}\mathrm{C}_1+{}^{100}\mathrm{C}_2..........-{}^{100}\mathrm{C}_{49}$
$=\frac{-1}{2}{}^{100}\mathrm{C}_{50}$
$=\frac{-1}{2}\left(\frac{100}{50}\right){}^{99}\mathrm{C}_{49}$

$=-{}^{99}\mathrm{C}_{49}$