First slide

Binomial theorem for positive integral Index

Question

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

Moderate

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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