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Binomial Theorem question

2002 · Shift 0 · Q106

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JEE Main
Mathematics
Binomial Theorem
2002 · Shift 0 · Q106

Binomial Theorem question

2002 · Shift 0 · Q106

JEE MainMathematicsBinomial TheoremMCQ+4 / −1

The positive integer just greater than

{\left( {1 + 0.0001} \right)^{10000}}

is

A
4
B
5
C
2
D
3

View written solutionFree

Correct answer: D

Let $x=\left(1+0.0001\right)^{10000}=\left(1+\frac{1}{10000}\right)^{10000}.$

We need the positive integer just greater than $x$ , i.e. the smallest integer greater than $x$ .

Compare $x$ with the known limit: $\left(1+\frac{1}{n}\right)^n<e \quad \text{for all positive integers } n.$ So, $x=\left(1+\frac{1}{10000}\right)^{10000}<e.$ Since $e\approx 2.71828<3,$ we get $x<3.$
Now show that $x>2$ . Using the binomial theorem, $\left(1+\frac{1}{10000}\right)^{10000}=1+\binom{10000}{1}\frac{1}{10000}+\binom{10000}{2}\frac{1}{10000^2}+\cdots$ The first two terms give $1+\binom{10000}{1}\frac{1}{10000}=1+10000\cdot \frac{1}{10000}=2.$ All remaining terms are positive, hence $x>2.$
Therefore, $2<x<3.$ So the smallest integer greater than $x$ is $3.$
Checking options:

A: $4$ — incorrect
B: $5$ — incorrect
C: $2$ — incorrect, since number is greater than $2$
D: $3$ — correct

Hence the correct answer is D.

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