Sequences and Series 4 Question 2
2. Let $S _n=1+q+q^{2}+\ldots+q^{n}$ and $T _n=1+\frac{q+1}{2}+\frac{q+1}{2}^{2}+\ldots+\frac{q+1}{2}^{n}$, where $q$ is a real number and $q \neq 1$. If
${ }^{101} C _1+{ }^{101} C _2 \cdot S _1+\ldots+{ }^{101} C _{101} \cdot S _{100}=\alpha T _{100}$, then $\alpha$ is equal to
(a) $2^{100}$
(b) 202
(c) 200
(d) $2^{99}$
(2019 Main, 11 Jan II)
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Solution:
- (a) We have, $S _n=1+q+q^{2}+\ldots+q^{n}$ and
$T _n=1+\frac{q+1}{2}+\frac{q+1}{2}^{2}+\ldots+\frac{q+1}{2}^{n}$
Also, we have
$$ \begin{aligned} & { }^{101} C _1+{ }^{101} C _2 S _1+{ }^{101} C _3 S _2+\ldots+{ }^{101} C _{101} S _{100}=\alpha T _{100} \\ & \Rightarrow{ }^{101} C _1+{ }^{101} C _2(1+q)+{ }^{101} C _3\left(1+q+q^{2}\right) \\ & +\ldots+{ }^{101} C _{101}\left(1+q+q^{2}+\ldots+q^{100}\right) \\ & =\alpha \cdot T _{100} \\ & \Rightarrow{ }^{101} C _1+{ }^{101} C _2 \frac{\left(1-q^{2}\right)}{1-q}+{ }^{101} C _3 \frac{1-q^{3}}{1-q} \\ & +{ }^{101} C _4 \frac{1-q^{4}}{1-q}+\ldots+{ }^{101} C _{101} \frac{1-q^{101}}{1-q} \\ & =\alpha \cdot T _{100} \quad\left[\because \text { for a GP }, S _n=a \frac{1-r^{n}}{1-r}, r \neq 1\right] \\ & \Rightarrow \quad \frac{1}{1-q}\left[{{ }^{101} C _1+{ }^{101} C _2+\ldots+{ }^{101} C _{101} }\right. \\ & -{{ }^{101} C _1 q+{ }^{101} C _2 q^{2}+\ldots+{ }^{101} C _{101} q^{101} }=\alpha \cdot T _{100} \\ & \Rightarrow \quad \frac{1}{(1-q)}\left[\left(2^{101}-1\right)-\left((1+q)^{101}-1\right)\right]=\alpha T _{100} \end{aligned} $$
$$ \begin{aligned} & {\left[\because q \neq 1 \Rightarrow q+1 \neq 2 \Rightarrow \frac{q+1}{2} \neq 1\right]} \\ & =\frac{\alpha\left[2^{101}-(q+1)^{101}\right]}{(1-q) \cdot 2^{100}} \Rightarrow \alpha=2^{100} \end{aligned} $$
