Use the identity $\frac{{}^n{C}_r}{r+1}=\frac{1}{n+1}\cdot {}^{n+1}{C}_{r+1}$:

$\frac{{}^{100}{C}_r}{r+1}=\frac{1}{101}\cdot {}^{101}{C}_{r+1}$

Thus the sum becomes:

$S=\frac{1}{101}{\sum }_{r=50}^{100}{}^{101}{C}_{r+1}=\frac{1}{101}{\sum }_{k=51}^{101}{}^{101}{C}_k$

By symmetry ${\sum }_{k=0}^{101}{}^{101}{C}_k=2^{101}$, and the upper half ${\sum }_{k=51}^{101}=\frac{2^{101}}{2}=2^{100}$.

$\therefore S=\frac{2^{100}}{101}$.