STD 11 - 7. binomial theoram — JEE STD 11 Science — Question

Let $x=10^{100}$

$\Rightarrow P=\left(1+\frac{1}{x}\right)^{x}$

$\Rightarrow P=1+(x)\left(\frac{1}{x}\right)+\frac{(x)(x-1)}{\lfloor 2} \cdot \frac{1}{x^{2}}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+\frac{(x)(x-1)(x-2)}{\lfloor 3} \cdot \frac{1}{x^{3}}+\ldots$

(upto $10^{100}+1$ terms $)$

$\Rightarrow P=1+1+\left(\frac{1}{\lfloor2}-\frac{1}{\lfloor2 x^{2}}\right)+\left(\frac{1}{\lfloor3}-\ldots\right)+\ldots \text { so on }$

$\Rightarrow P=2+\left(\text { Positive value less than } \frac{1}{\lfloor 2}+\frac{1}{\lfloor 3}+\frac{1}{\lfloor 4}+\ldots\right)$

$\Rightarrow P \in(2,3)$

Also $e=1+\frac{1}{\lfloor1}+\frac{1}{\lfloor2}+\frac{1}{\lfloor3}+\frac{1}{\lfloor4}+\ldots$

$\Rightarrow \frac{1}{\lfloor2}+\frac{1}{\lfloor 3}+\frac{1}{\lfloor 4}+\ldots=\mathrm{e}-2$

$\Rightarrow \mathrm{P}=2+($ Positive value less than $\mathrm{e}-2)$

$\Rightarrow \mathrm{P} \in(2,3)$

$\Rightarrow$ least integer value of $\mathrm{P}$ is $3$