Let $S_{n}=1+q+q^{2}+\ldots .+q^{n}$ and…

Let $S_{n}=1+q+q^{2}+\ldots .+q^{n}$ and $\mathrm{T}_{\mathrm{n}}=1+\left(\frac{\mathrm{q}+1}{2}\right)+\left(\frac{\mathrm{q}+1}{2}\right)^{2}+\ldots+\left(\frac{\mathrm{q}+1}{2}\right)^{\mathrm{n}}$ where $q$ is a real number and $q \neq 1$. If ${ }^{101} \mathrm{C}_{1}+{ }^{101} \mathrm{C}_{2} \cdot \mathrm{S}_{1}+\ldots .+{ }^{101} \mathrm{C}_{101} \cdot \mathrm{S}_{100}=\alpha \mathrm{T}_{100},$ then $\alpha$ is equal to :
  1. $2^{99}$
  2. 202
  3. 200
  4. $2^{100}$

Solution

$S_{n}=\left(\frac{1-q^{n+1}}{1-q}\right), T_{n}=\frac{1-\left(\frac{q+1}{2}\right)^{n+1}}{1-\left(\frac{q+1}{2}\right)}$ $\Rightarrow T_{100}=\frac{1-\left(\frac{q+1}{2}\right)^{101}}{1-\left(\frac{q+1}{2}\right)}$ $S n=\frac{1}{1-q}-\frac{q^{n+1}}{1-q}, T_{100}=\frac{2^{101}-(\mathrm{q}+1)^{101}}{2^{100}(1-\mathrm{q})}$ Now, ${ }^{101} C_{1}+{ }^{101} C_{2} S_{1}+{ }^{101} C_{3} S_{2}+\ldots+{ }^{101} C_{101} S_{100}$ $=\left(\frac{1}{1-q}\right)\left({ }^{101} C_{2}+\ldots+{ }^{101} C_{101}\right)$ $ \begin{aligned} & \quad-\frac{1}{1-q}\left({ }^{101} C_{2} q^{2}+{ }^{101} C_{3} q^{3}+\ldots+{ }^{101} C_{101} q^{101}\right)+101 \\ =& \frac{1}{1-q}\left(2^{101}-1-101\right)-\left(\frac{1}{1-q}\right)\left((1+q)^{101}-1\right.\\ \left.-{ }^{101} C_{1} q\right)+101 \\ =& \frac{1}{1-q}\left[2^{101}-102-(1+q)^{101}+1+101 q\right]+101 \\ =& \frac{1}{1-q}\left[2^{101}-101+101 q-(1+q)^{101}\right]+101 \\ =&\left(\frac{1}{1-q}\right)\left[2^{101}-(1+q)^{101}\right]=2^{100} T_{100} \end{aligned} $ Hence, by comparison $\alpha=2^{100}$

Asked in: JEE Main 2019 (11 Jan Shift 2)

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