If the solve the problem

Question:

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 :-

  1. $2^{100}$

  2. 200

  3. $2^{99}$

  4. 202


Correct Option: 1

Solution:

${ }^{101} \mathrm{C}_{1}+{ }^{101} \mathrm{C}_{2} \mathrm{~S}_{1}+\ldots \ldots+{ }^{101} \mathrm{C}_{101} \mathrm{~S}_{100}$

$=\alpha \mathrm{T}_{100}$

${ }^{101} \mathrm{C}_{1}+{ }^{101} \mathrm{C}_{2}(1+q)+{ }^{101} \mathrm{C}_{3}\left(1+q+\mathrm{q}^{2}\right)+$

$\ldots \ldots+{ }^{101} \mathrm{C}_{101}\left(1+q+\ldots \ldots+\mathrm{q}^{100}\right)$

$=2 \alpha \frac{\left(1-\left(\frac{1+q}{2}\right)^{101}\right)}{(1-q)}$

$\Rightarrow{ }^{101} C_{1}(1-q)+{ }^{101} C_{2}\left(1-q^{2}\right)+$

$\ldots .+{ }^{101} C_{101}\left(1-q^{101}\right)$

$=2 \alpha\left(1-\left(\frac{1+q}{2}\right)^{101}\right)$

$\Rightarrow\left(2^{101}-1\right)-\left((1+q)^{101}-1\right)$

$=2 \alpha\left(1-\left(\frac{1+q}{2}\right)^{101}\right)$

$\Rightarrow 2^{101}\left(1-\left(\frac{1+q}{2}\right)^{101}\right)=2 \alpha\left(1-\left(\frac{1+q}{2}\right)^{101}\right)$

$\Rightarrow \alpha=2^{100}$

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