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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 :
Options:
$\mathrm{T}_{\mathrm{n}}=1+\left(\frac{\mathrm{q}+1}{2}\right)+\left(\frac{\mathrm{q}+1}{2}\right)^{2}+\ldots+\left(\frac{\mathrm{q}+1}{2}\right)^{\mathrm{n}}$
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 :
Solution:
1600 Upvotes
Verified Answer
The correct answer is:
$2^{100}$
$S_{n}=\left(\frac{1-q^{n+1}}{1-q}\right), T_{n}=\frac{1-\left(\frac{q+1}{2}\right)^{n+1}}{1-\left(\frac{q+1}{2}\right)}$
$\Rightarrow T_{100}=\frac{1-\left(\frac{q+1}{2}\right)^{101}}{1-\left(\frac{q+1}{2}\right)}$
$S n=\frac{1}{1-q}-\frac{q^{n+1}}{1-q}, T_{100}=\frac{2^{101}-(\mathrm{q}+1)^{101}}{2^{100}(1-\mathrm{q})}$
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 $\alpha=2^{100}$
$\Rightarrow T_{100}=\frac{1-\left(\frac{q+1}{2}\right)^{101}}{1-\left(\frac{q+1}{2}\right)}$
$S n=\frac{1}{1-q}-\frac{q^{n+1}}{1-q}, T_{100}=\frac{2^{101}-(\mathrm{q}+1)^{101}}{2^{100}(1-\mathrm{q})}$
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 $\alpha=2^{100}$
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