Set $S_n=1+q+q^2+\dots +q^n$ and $T_n=1+\left(\frac{q+1}{2}\right)+{\left(\frac{q+1}{2}\right)}^2+\dots +{\left(\frac{q+1}{2}\right)}^n$where $q$ is a real number and $q\ne 1$ ${If }^{101}{C}_1{+}^{101}{C}_2{S}_1+\dots {+}^{101}$ $C_{101}{S}_{100}=\alpha T_{100}$ , then $\alpha$ is equal to 

Solution:

Here, ${ }^{101}{C}_1{+}^{101}{C}_2{S}_1+\dots +101$$C_{101}{S}_{100}=\alpha T_{100}$

${\Rightarrow }^{101}{C}_1{+}^{101}{C}_2(1+q){+}^{101}{C}_3\left(1+q+q^2\right)+\dots$${+}^{101}{C}_{101}\left(1+q+\dots +q^{100}\right)$

$\begin{array}{l}=2\alpha \frac{\left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)}{(1-q)}\\ {\Rightarrow }^{101}{C}_1(1-q){+}^{101}{C}_2\left(1-q^2\right)+\dots {+}^{101}{C}_{101}\left(1-q^{101}\right)\\ =2\alpha \left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)\\ \Rightarrow \left(2^{101}-1\right)-\left[(1+q{)}^{101}-1\right]=2\alpha \left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)\\ \Rightarrow 2^{101}\left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)=2\alpha \left(1-{\left(\frac{1+q}{2}\right)}^{101}\right)\Rightarrow \alpha =2^{100}\end{array}$