Q. If $A = 1000^{1000}$ and $B = {(1001)}^{999}$, then
A) $A > B$
B) $A = B$
C) $A < B$
D) None of these
Explanation
(a)
Since, $\left( 1 + \frac{1}{n} \right)^{n} < 3\ for\ \forall n \in N$
Now, $\frac{{(1001)}^{999}}{{(1000)}^{1000}} = \frac{1}{1001} \bullet \left( \frac{1001}{1000} \right)^{1000}$
$= \frac{1}{1001}\left( 1 + \frac{1}{1000} \right)^{1000} < \frac{1}{1001} \bullet 3 < 1$
${(1001)}^{999} < {(1000)}^{1000}$
$\therefore\ B < A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.