Using the Binomial Theorem,determine which number is larger: $(1.1)^{10000}$ or $1000$.
(A) Using the Binomial Theorem,we expand $(1.1)^{10000}$ as follows:
$(1.1)^{10000} = (1 + 0.1)^{10000}$
$= {^{10000}}C_0 (1)^{10000} + {^{10000}}C_1 (1)^{9999} (0.1) + \text{other positive terms}$
$= 1 + 10000 \times 0.1 + \text{other positive terms}$
$= 1 + 1000 + \text{other positive terms}$
$= 1001 + \text{other positive terms}$
Since all terms in the expansion are positive,it is clear that $1001 + \text{other positive terms} > 1000$.
Therefore,$(1.1)^{10000} > 1000$.
$(1.1)^{10000} = (1 + 0.1)^{10000}$
$= {^{10000}}C_0 (1)^{10000} + {^{10000}}C_1 (1)^{9999} (0.1) + \text{other positive terms}$
$= 1 + 10000 \times 0.1 + \text{other positive terms}$
$= 1 + 1000 + \text{other positive terms}$
$= 1001 + \text{other positive terms}$
Since all terms in the expansion are positive,it is clear that $1001 + \text{other positive terms} > 1000$.
Therefore,$(1.1)^{10000} > 1000$.
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