Let $s_n = 1 + q + q^2 + ................. + q^n$ and

$T_n = 1 + \left(\dfrac{q + 1}{2}\right) + \left(\dfrac{q + 1}{2}\right)^2 +......... \left(\dfrac{q + 1}{2}\right)^n$

where q is a real number and q $\neq$ 1.

If $^{101}C_1 + ^{101}{C_2}.S_{1} +...... + ^{101}C_{101}.S_{100} = \alpha T_{100}, then\,\,\, \alpha$ is equal to :-

$\begin{array}{l} S_{n}=1+q+q^{2}+\cdots+q^{n} \\ \Rightarrow S_{n}=\frac{1-q^{n+1}}{1-q} \cdots(1) \\ T_{n}=1+\left(\frac{q+1}{2}\right)+\left(\frac{q+1}{2}\right)+\cdots+\left(\frac{q-1}{2}\right)^{n} \\ \Rightarrow T_{n}=\frac{1-\left(\frac{q+1}{2}\right)^{n+1}}{1-\left(\frac{q+1}{2}\right)} \\ \Rightarrow T_{n}=\frac{2^{n+1}-(q+1)^{n+1}}{2^{n} \cdot(1-q)} \\ \Rightarrow T_{100}=\frac{2^{101}-(q+1)^{101}}{2^{100}(1-q)} …(2)\end{array}$

Now, ${ }^{101} \mathrm{C}_{1}+{ }^{101} \mathrm{C}_{2} \cdot \mathrm{S}_{1}+\cdots+{ }^{101} \mathrm{C}_{101} \cdot \mathrm{S}_{100}=\mathrm{a} \mathrm{T}_{100}$

$\Rightarrow \sum_{r=1}^{101}{ }^{101} C_{r} S_{r-1}=a T_{100}$

From equation (1)

$\begin{array}{l} S_{r-1}=\frac{1-q^{r}}{1-q} \\ \Rightarrow \sum_{r=1}^{101}{ }^{101} C_{r}\left(\frac{1-q^{r}}{1-q}\right)=a T_{100} \\ \Rightarrow \frac{1}{1-q}\left[\sum_{r=1}^{101}{ }^{101} C_{r}-\sum_{r=1}^{101}{ }^{101} C_{r} \cdot q^{r}\right]=a T_{100} \\ \Rightarrow \frac{1}{1-q}\left[\left(2^{101}-1\right)-\left((1+q)^{101}-1\right)\right]=a T_{100} \end{array}$

From equation (2), we get

$\begin{array}{l} \frac{1}{1-q}\left[2^{101}-(1+q)^{101}\right]=a \cdot\left[\frac{2^{101}-(1+q)^{10}}{2^{100}(1-q)}\right] \\ \Rightarrow a=2^{100} \end{array}$