Question

The lowest integer which is greater than ${\left(1+\frac{1}{10^{100}}\right)}^{10^{100}}$ is_______.

• A. 3
• B. 4
• C. 2
• D. 1

Answer 1

Answer: (A)

Let $P={\left(1+\frac{1}{10^{100}}\right)}^{10^{100}}$
Let $x=10^{100}$
$\Rightarrow P={\left(1+\frac{1}{x}\right)}^x$
$\Rightarrow P=1+(x)\left(\frac{1}{x}\right)+\frac{(x)(x-1)}{2!}\cdot \frac{1}{x^2}+\frac{(x)(x-1)(x-2)}{3!}\cdot \frac{1}{x^3}+\dots \dots \dots$ (upto $10^{100}+1$ terms )
$\Rightarrow P=1+1+\left(\frac{1}{2!}-\frac{1}{2!x^2}\right)+\left(\frac{1}{3!}-\dots \dots \right)+$ so on
$\Rightarrow P=2+($ Positive value less then $\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots .)$
Also e $=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots \dots$
$\Rightarrow \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots .=e-2$
$\Rightarrow P=2+($ positive value less then $e-2)$
$\Rightarrow P\in (2,3)$
$\Rightarrow$ least integer value of $P$ is $3$