If A=10001000 and B=(1001)999 then

If $A = 1000^{1000}$ and $B = (1001)^{999}$ then

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

Solution:

Since, ${(1 + \frac{1}{n})}^{n} < 3 for \forall n \in N$

Now $\frac{(1001)^{999}}{(1000)^{1000}} = \frac{1}{1001} \cdot {(\frac{1001}{1000})}^{1000}$

$= \frac{1}{1001} {(1 + \frac{1}{1000})}^{1000} < \frac{1}{1001} \cdot 3 < 1$

$(1001)^{999} < (1000)^{1000}$

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