Set $S_{n} = 1 + q + q^{2} + \dots + q^{n}$ and $T_{n} = 1 + (\frac{q + 1}{2}) + {(\frac{q + 1}{2})}^{2} + \dots + {(\frac{q + 1}{2})}^{n}$ where $q$ is a real number and $q \neq 1$ ${I f}^{101} C_{1} +^{101} C_{2} S_{1} + \dots +^{101}$ $C_{101} S_{100} = α T_{100}$ , then $α$ is equal to

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$2^{99}$

202

200

$2^{100}$

answer is A.

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Detailed Solution

Here, $^{101} C_{1} +^{101} C_{2} S_{1} + \dots + 101$ $C_{101} S_{100} = α T_{100}$

$\Rightarrow^{101} C_{1} +^{101} C_{2} (1 + q) +^{101} C_{3} (1 + q + q^{2}) + \dots$ $+^{101} C_{101} (1 + q + \dots + q^{100})$

$\begin{array}{l} = 2 α \frac{(1 - {(\frac{1 + q}{2})}^{101})}{(1 - q)} \\ \Rightarrow^{101} C_{1} (1 - q) +^{101} C_{2} (1 - q^{2}) + \dots +^{101} C_{101} (1 - q^{101}) \\ = 2 α (1 - {(\frac{1 + q}{2})}^{101}) \\ \Rightarrow (2^{101} - 1) - [(1 + q)^{101} - 1] = 2 α (1 - {(\frac{1 + q}{2})}^{101}) \\ \Rightarrow 2^{101} (1 - {(\frac{1 + q}{2})}^{101}) = 2 α (1 - {(\frac{1 + q}{2})}^{101}) \Rightarrow α = 2^{100} \end{array}$