First slide

Binomial theorem for positive integral Index

Question

The positive integer just greater than $S = {(1 + 0.0001)}^{10000}$ is

Moderate

A

2
B

3
C

4
D

5

Solution

Let $n = 10000,$ then
$\begin{array}{l} S = {(1 + \frac{1}{n})}^{n} \\ = 1 +^{n} C_{1} (\frac{1}{n}) +^{n} C_{2} {(\frac{1}{n})}^{2} + \dots +^{n} C_{n} {(\frac{1}{n})}^{n} \\ = 1 + 1 + \frac{1}{2!} \frac{n (n - 1)}{n^{2}} + \frac{1}{3!} \frac{n (n - 1) (n - 2)}{n^{3}} + \dots + \frac{1}{n!} \frac{n (n - 1) \dots (n - n + 1)}{n^{n}} \\ = 1 + 1 + \frac{1}{2!} (1 - \frac{1}{n}) + \frac{1}{3!} (1 - \frac{1}{n}) (1 - \frac{2}{n}) + \dots + \frac{1}{n!} (1 - \frac{1}{n}) \dots (1 - \frac{n - 1}{n}) \\ < 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!} \end{array}$
But $\frac{1}{r!} = \frac{1}{1.2 \dots r} \leq \frac{1}{2^{r - 1}} \forall r \geq 2 .$
Thus, $S \leq 1 + 1 + \frac{1}{2} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{n - 1}}$
$= 1 + \frac{1 - (1 / 2)^{n}}{1 - 1 / 2} = 3 - {(\frac{1}{2})}^{n - 1} < 3 .$

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