Question

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

• A. $1$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is C $3$

Let ${10}^{100}=n$
Then, ${(1+\frac{1}{n})}^n$
$={}^n{C}_0+{}^n{C}_1\cdot \frac{1}{n}+{}^n{C}_2\cdot \frac{1}{n^2}+{}^n{C}_3\cdot \frac{1}{n^3}+\cdots$
$=1+1+\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{6n^3}+\cdots$
$\Rightarrow {(1+\frac{1}{n})}^n>2$
We know that
$\underset{n\to \infty }{lim}{(1+\frac{1}{n})}^n=e\approx 2.72$
Hence, lowest integer is $3$ which is greater than ${(1+\frac{1}{{10}^{100}})}^{{10}^{100}}$