Question

Let $a=\displaystyle \sum_{n=101}^{200} 2^{n} \sum_{k=101}^{n} \frac{1}{k !}$ and $b =\displaystyle \sum_{ n =101}^{200} \frac{2^{201}-2^{ n }}{ n !}$ Then $\frac{ a }{ b }$ is

• A. 1
• B. $\frac{3}{2}$
• C. 2
• D. $\frac{5}{2}$

Answer 1

Answer: (A)

$a =\displaystyle\sum_{ n =101}^{200} 2^{ n } \displaystyle\sum_{ k =101}^{ n } \frac{1}{ k !}$
$=\frac{2^{101}}{101 !}+2^{1 m }\left(\frac{1}{101 !}+\frac{1}{102 !}\right)+2^{103}\left(\frac{1}{101 !}+\frac{1}{102 !}+\frac{1}{103 !}\right)+\ldots$
$+2^{200}\left(\frac{1}{101 !}+\frac{1}{102 !}+\ldots+\frac{1}{200 !}\right)$
$\frac{2^{101}+\ldots+2^{200}}{101 !}+\frac{2^{102}+\ldots+2^{200}}{102 !}+\ldots+\frac{2^{200}}{200 !}$
$=\frac{2^{101}\left(2^{100}-1\right)}{101 !}, \frac{2^{102}\left(2^{99}-1\right)}{102 !}, \ldots 1 \frac{2^{200}}{200 !}$
$=\left(\frac{2^{201}}{101 !}-\frac{2^{101}}{1011}\right)+\left(\frac{2^{201}}{102 !}-\frac{2^{102}}{102 !}\right)+\ldots+\left(\frac{2^{201}}{200 !}-\frac{2^{200}}{200 !}\right)$
$=\displaystyle\sum_{ n =101}^{200} \frac{2^{201}-2^{ n }}{ n !}= b$
$\therefore \frac{ a }{ b }= 1$