Question

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

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

Answer 1

Answer: (A)

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