Let a=∑n=1012002^n∑k=101n1/(k!) and b=∑n=101200(2^201-2^n)/(n!) Then a/b is

Correct answer: (A) 1. a=∑n=1012002^n∑k=101n1/(k!) =2^101101!+2^1021/(101!)+1/(102!)+2^1031/(101!)+1/(102!)+1/(103!)+… +2^2001/(101!)+1/(102!)+…+1/(200!) (2^101+…+2^200)/(101!)+(2^102+…+2^200)/(102!)+…+2^200200! =(2^1012^100-1)/(101!)+(2^1022^99-1)/(102!)+…+2^200200! =2^201101!-2^101101!+2^201102!-2^102102!+…+2^201200!-2^200200! =∑n=101200(2^201-2^n)/(n!)=b ∴a/b=1

Q57 / 91 · KVPY 2020 (SB/SX) · Mathematics

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

Correct answer
A
$1$
B
$\frac{3}{2}$
C
$2$
D
$\frac{5}{2}$

Solution

$a = \sum_{n = 101}^{200} 2^{n} \sum_{k = 101}^{n} \frac{1}{k!}$

$= \frac{2^{101}}{101!} + 2^{102} (\frac{1}{101!} + \frac{1}{102!}) + 2^{103} (\frac{1}{101!} + \frac{1}{102!} + \frac{1}{103!}) + \dots$

$+ 2^{200} (\frac{1}{101!} + \frac{1}{102!} + \dots + \frac{1}{200!})$

$\frac{2^{101} + \dots + 2^{200}}{101!} + \frac{2^{102} + \dots + 2^{200}}{102!} + \dots + \frac{2^{200}}{200!}$

$= \frac{2^{101} (2^{100} - 1)}{101!} + \frac{2^{102} (2^{99} - 1)}{102!} + \dots + \frac{2^{200}}{200!}$

$= (\frac{2^{201}}{101!} - \frac{2^{101}}{101!}) + (\frac{2^{201}}{102!} - \frac{2^{102}}{102!}) + \dots + (\frac{2^{201}}{200!} - \frac{2^{200}}{200!})$

$= \sum_{n = 101}^{200} \frac{2^{201} - 2^{n}}{n!} = b$

$∴ \frac{a}{b} = 1$

Hence, the correct answer is option: 'A'.

Let a=∑n=1012002n∑k=101n1k! and b=∑n=1012002201-2nn! Then ab is

Solution

More Sequences and Series Questions

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