Question

Let $S_n + 1 + q + q^2 + ....... + q^n $ and $T_{n} = 1+ \left(\frac{q+1}{2}\right) + \left(\frac{q+1}{2}\right)^{2} +..... +\left(\frac{q+1}{2}\right)^{n} $ where q is a real number and $q \neq 1$.
If ${^{101}C_1} + {^{101}C_2} .S_1 + ...... + {^{101}C_{101}}.S_{100} = \alpha T_{100}$, then $\alpha$ is equal to :-

• A. $2^{100}$
• B. 200
• C. $2^{99}$
• D. 202

Answer 1

Answer: (A)

$^{101}C_{1} +^{101}C_{2}S_{1} +.....+^{101}C_{101} S_{100} = \alpha T_{100} $
$^{101}C_{1} +^{101}C_{2}\left(1+q\right) +^{101}C_{3}\left(1+q+q^{2}\right) +.... +^{101}C_{101} \left(1+q+....+q^{100}\right) $
$ = 2\alpha \frac{\left(1- \left(\frac{1+q}{2}\right)^{101}\right)}{\left(1-q\right)} $
$\Rightarrow ^{101}C_{1}\left(1-q\right)+^{101}C_{2}\left(1-q^{2}\right)+.....+^{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(\left(1+q\right)^{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} $