Let $\mathrm{S}_{\mathrm{n}}=1+\mathrm{q}+\mathrm{q}^{2}+\ldots \ldots .+\mathrm{q}^{\mathrm{n}}$ and
$\mathrm{T}_{\mathrm{n}}=1+\left(\frac{\mathrm{q}+1}{2}\right)+\left(\frac{\mathrm{q}+1}{2}\right)^{2}+\ldots \ldots+\left(\frac{\mathrm{q}+1}{2}\right)^{\mathrm{n}}$
where $\mathrm{q}$ is a real number and $\mathrm{q} \neq 1$
If ${ }^{101} \mathrm{C}_{1}+{ }^{101} \mathrm{C}_{2} \cdot \mathrm{S}_{1}+\ldots \ldots+{ }^{101} \mathrm{C}_{101} \cdot \mathrm{S}_{100}=\alpha \mathrm{T}_{100}$. then $\alpha$ is equal to :-
Correct Option: 1
${ }^{101} \mathrm{C}_{1}+{ }^{101} \mathrm{C}_{2} \mathrm{~S}_{1}+\ldots \ldots+{ }^{101} \mathrm{C}_{101} \mathrm{~S}_{100}$
$=\alpha \mathrm{T}_{100}$
${ }^{101} \mathrm{C}_{1}+{ }^{101} \mathrm{C}_{2}(1+q)+{ }^{101} \mathrm{C}_{3}\left(1+q+\mathrm{q}^{2}\right)+$
$\ldots \ldots+{ }^{101} \mathrm{C}_{101}\left(1+q+\ldots \ldots+\mathrm{q}^{100}\right)$
$=2 \alpha \frac{\left(1-\left(\frac{1+q}{2}\right)^{101}\right)}{(1-q)}$
$\Rightarrow{ }^{101} C_{1}(1-q)+{ }^{101} C_{2}\left(1-q^{2}\right)+$
$\ldots .+{ }^{101} C_{101}\left(1-q^{101}\right)$
$=2 \alpha\left(1-\left(\frac{1+q}{2}\right)^{101}\right)$
$\Rightarrow\left(2^{101}-1\right)-\left((1+q)^{101}-1\right)$
$=2 \alpha\left(1-\left(\frac{1+q}{2}\right)^{101}\right)$
$\Rightarrow 2^{101}\left(1-\left(\frac{1+q}{2}\right)^{101}\right)=2 \alpha\left(1-\left(\frac{1+q}{2}\right)^{101}\right)$
$\Rightarrow \alpha=2^{100}$