Key Concepts and Formulas

This problem leverages our understanding of the mathematical constant $e$, specifically its definition through limits and infinite series, in conjunction with the Binomial Theorem.

1. The Binomial Theorem: For any non-negative integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ A specific form useful here is for $(1+x)^n$: $(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n$

2. The Constant $e$ (Euler's Number):

   • Limit Definition: $e$ is defined as the limit of a sequence: $e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$
   • Series Expansion: $e$ can also be represented by an infinite series: $e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ The approximate value of $e$ is $2.71828...$.

3. Crucial Inequality: For any finite positive integer $n$, the value of $\left(1 + \frac{1}{n}\right)^n$ is always strictly less than $e$. That is, $\left(1 + \frac{1}{n}\right)^n < e$. This inequality arises from comparing the finite binomial expansion with the infinite series for $e$.

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Step-by-Step Solution

Step 1: Transform the Given Expression into a Standard Form

The given expression is ${\left( {1 + 0.0001} \right)^{10000}}$.

• Why this step? Our goal is to relate this expression to the constant $e$. The first step is to rewrite the decimal $0.0001$ as a fraction, which will reveal a form directly comparable to the definition of $e$.
• We can write $0.0001$ as $\frac{1}{10000}$.
• Let $n = 10000$. Substituting this into the expression, we get: ${\left( {1 + \frac{1}{n}} \right)^n}$ This is precisely the form that appears in the limit definition of $e$.

Step 2: Expand the Expression Using the Binomial Theorem

Now, we will apply the Binomial Theorem to expand ${\left( {1 + \frac{1}{n}} \right)^n}$ for $n=10000$.

• Why this step? Expanding the expression allows us to examine its individual terms and compare them with the terms of the series expansion for $e$. This term-by-term comparison is essential to establish the inequality between our expression and $e$.
• Using the Binomial Theorem for $(1+x)^n$ with $x = \frac{1}{n}$: ${\left( {1 + \frac{1}{n}} \right)^n} = \binom{n}{0}(1)^n\left(\frac{1}{n}\right)^0 + \binom{n}{1}(1)^{n-1}\left(\frac{1}{n}\right)^1 + \binom{n}{2}(1)^{n-2}\left(\frac{1}{n}\right)^2 + \binom{n}{3}(1)^{n-3}\left(\frac{1}{n}\right)^3 + \dots + \binom{n}{n}(1)^0\left(\frac{1}{n}\right)^n$
• Let's simplify each term:
  • $\binom{n}{0} = 1$
  • $\binom{n}{1}\left(\frac{1}{n}\right) = n \cdot \frac{1}{n} = 1$
  • $\binom{n}{2}\left(\frac{1}{n}\right)^2 = \frac{n(n-1)}{2!} \cdot \frac{1}{n^2} = \frac{1}{2!} \cdot \frac{n(n-1)}{n^2} = \frac{1}{2!} \left(1 - \frac{1}{n}\right)$
  • $\binom{n}{3}\left(\frac{1}{n}\right)^3 = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{n^3} = \frac{1}{3!} \cdot \frac{n(n-1)(n-2)}{n^3} = \frac{1}{3!} \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)$
• Continuing this pattern, the full expansion becomes: ${\left( {1 + \frac{1}{n}} \right)^n} = 1 + 1 + \frac{1}{2!} \left(1 - \frac{1}{n}\right) + \frac{1}{3!} \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \dots + \frac{1}{n!} \left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\dots\left(1 - \frac{n-1}{n}\right)$ Note that this is a finite series with $(n+1)$ terms.

Step 3: Compare with the Series Expansion of $e$

Now, let's compare the binomial expansion we just derived with the infinite series for $e$: $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!} + \frac{1}{(n+1)!} + \dots$ Rewriting the first two terms:

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