Using Binomial Theorem, indicate which number is large $(1.1)^{10000}$ or $1000$ .

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Solution

$(1.1)^{10000} = (1 + .1)^{1000} = 1 +^{10000} C_{1} (0.1) +^{10000} C_{2} (.1)^{2} + . . . . . +^{1000} C_{10000} (.1)^{10000}$
$= 1 + 10000 (.1) +^{10000} C_{2} (.1)^{2} + . . . . . +^{1000} C_{10000} (.1)^{10000}$
$= 1 + 1000 +^{10000} C_{2} (.1)^{2} + . . . . . +^{1000} C_{10000} (.1)^{10000} > 1000$
Hence $(1.1)^{10000}$ is greater than $1000$

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