Question

if $A={1000}^{1000}$ and $B={\left(1001\right)}^{999}$, then

• A. $A>B$
• B. $A=B$
• C. $A<B$
• D. None of these

Answer 1

Answer: (A)

$A>B$

$A={1000}^{1000}$

$B={\left(1001\right)}^{999}$

$\frac{B}{A}=\frac{{\left(1001\right)}^{999}}{{1000}^{1000}}$

$\frac{{\left(1001\right)}^{999}}{{\left(1000\right)}^{1000}}\times \frac{1001}{1001}$

$\frac{{\left(1001\right)}^{1000}}{{\left(1000\right)}^{1000}}\times \frac{1}{1001}$

$\frac{1}{1001}\times {\left(\frac{1001}{1000}\right)}^{1000}$

$\frac{1}{1001}{\left(\frac{1000+1}{1000}\right)}^{1000}$

$\frac{1}{1001}{\left(1+\frac{1}{1000}\right)}^{1000}$

$Now\text{we know Lt}{\left(1+\frac{1}{n}\right)}^n=e\simeq 2.710$

$\therefore \frac{B}{A}\simeq \frac{1}{1001}\times e\text{< 1}$

B < A = A > B

${1000}^{1000}\text{<}{\left(1001\right)}^{999}$