Binomial theorem for positive integral Index

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

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Solution

For any $n \in N$ , we have
$\begin{array}{l} {(1 + \frac{1}{n})}^{n} = 1 + n \cdot \frac{1}{n} + \frac{n (n - 1)}{2!} {(\frac{1}{n})}^{2} + \frac{n (n - 1) (n - 2)}{3!} {(\frac{1}{n})}^{3} + \dots \\ \Rightarrow {(1 + \frac{1}{n})}^{n} = 1 + 1 + \frac{1}{2!} (1 - \frac{1}{n}) + \frac{1}{3!} (1 - \frac{1}{n}) (1 - \frac{2}{n}) + \dots \\ \Rightarrow {(1 + \frac{1}{n})}^{n} = 2 + \frac{1}{2!} (1 - \frac{1}{n}) + \frac{1}{3!} (1 - \frac{1}{n}) (1 - \frac{2}{n}) + \dots \end{array}$
$\Rightarrow {(1 + \frac{1}{n})}^{n} = 2 +$ a positive real number
$\Rightarrow {(1 + \frac{1}{n})}^{n} > 2$ for al $n \in N$
We know that
$lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = e \approx 2.71$
$\Rightarrow 2 < {(1 + \frac{1}{n})}^{n} < 3$ for all $n \in N$
$\Rightarrow 2 < (1 + 0.0001)^{10000} < 3$ $[Putting n = 10000]$
Hence, the greatest integer just greater than $(1 + 0.0001)^{10000}$ is $3$ .

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

The positive integer just greater than (1+0.0001)10000 , is

Ready to Test Your Skills?

free study material

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