JEE PYQ: Sequence And Series Question 65
Question 65 - 2019 (11 Jan Shift 2)
Let $S_n = 1 + q + q^2 + \ldots + q^n$ and $T_n = 1 + \left(\frac{q+1}{2}\right) + \left(\frac{q+1}{2}\right)^2 + \ldots + \left(\frac{q+1}{2}\right)^n$ where $q$ is a real number and $q \neq 1$. If $^{101}C_1 + ^{101}C_2 S_1 + \ldots + ^{101}C_{101} S_{100} = \alpha T_{100}$, then $\alpha$ is equal to:
(1) $2^{99}$
(2) 202
(3) 200
(4) $2^{100}$
Show Answer
Answer: (4)
Solution
$S_n = \frac{1 - q^{n+1}}{1 - q}$, $T_n = \frac{1 - \left(\frac{q+1}{2}\right)^{n+1}}{1 - \frac{q+1}{2}} = \frac{2^{101} - (q+1)^{101}}{2^{100}(1-q)}$
Now, $^{101}C_1 + ^{101}C_2 S_1 + \ldots + ^{101}C_{101} S_{100}$
$= \frac{1}{1-q}\left(^{101}C_1 + ^{101}C_2 + \ldots + ^{101}C_{101}\right) - \frac{1}{1-q}(^{101}C_1 q + ^{101}C_2 q^2 + \ldots + ^{101}C_{101} q^{101}) + 101$
$= \frac{1}{1-q}\left[2^{101} - 1 - 101\right] - \frac{1}{1-q}\left[(1+q)^{101} - 1 - 101q\right] + 101$
$= \frac{1}{1-q}\left[2^{101} - (1+q)^{101}\right] = 2^{100} T_{100}$
Hence, by comparison $\alpha = 2^{100}$
BETA
