Problem 49 - Algebra and Trigonometry

Question

Simplify using the Binomial Theorem. Show that \((1.01)^{100}>2 .\) [Hint: Note that \((1.01)^{100}=(1+0.01)^{100},\) and use the Binomial Theorem to show that the sum of the first two terms of the expansion is greater than \(2 . ]\)

Step-by-Step Solution

Answer

The expansion using the first two terms shows that \((1.01)^{100} > 2\). Additional terms increase the sum further.

Step 1: Understanding the Problem

We need to prove that \((1.01)^{100} > 2\).Given, \((1.01)^{100} = (1 + 0.01)^{100}\).We will use the Binomial Theorem to expand this expression.

Step 2: Recall the Binomial Theorem

The Binomial Theorem states that for any positive integer \(n\) and any numbers \(a\) and \(b\),\[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k.\]In this problem, \(a = 1\), \(b = 0.01\), and \(n = 100\).

Step 3: Expand the First Two Terms

When we expand \((1 + 0.01)^{100}\),we get the first two terms as follows:- The first term is \(\binom{100}{0} (1)^{100} (0.01)^0 = 1\).- The second term is \(\binom{100}{1} (1)^{99} (0.01)^1 = 100 \times 0.01 = 1\).

Step 4: Sum the First Two Terms

The sum of the first two terms is:\[1 + 1 = 2\].Since there are more terms in the expansion (all positive), \((1.01)^{100}\) is greater than \(2\).

Step 5: Conclude the Proof

Since the sum of the first two terms \((1 + 1) > 2\) and additional positive terms follow,we can conclude that \((1.01)^{100} > 2\).

Key Concepts

Binomial Expansion

The Binomial Expansion is a method used to expand expressions that are raised to any power. Imagine multiplying a binomial, like \((a + b)\), by itself multiple times. The Binomial Theorem simplifies this by providing a formula to calculate these products without manually multiplying out each term. The theorem states:

• \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

This might look intimidating, but let’s break it down:

• \(k\) is the term number starting from zero.
• \(\binom{n}{k}\) represents the "binomial coefficient," which can be calculated as \(\frac{n!}{k!(n-k)!}\).
• The powers of \(a\) decrease as \(k\) increases, while powers of \(b\) go up.

In the case of \((1.01)^{100}\), we replace \(a\) with 1 and \(b\) with 0.01, helping us to estimate and prove that \((1.01)^{100}\) is greater than 2 using just the first few terms, due to additional positive contributions from further terms.

Inequalities

In mathematics, inequalities are a way to express that one quantity is larger or smaller than another. They are not just about calculating an exact answer, but rather understanding how values relate. When we work with inequalities, we use signs like:

• \(>\) which means "greater than."
• < which means "less than."
• \(\geq\) and \(\leq\), which mean "greater than or equal to" and "less than or equal to," respectively.

In the context of our original problem, we needed to prove \((1.01)^{100} > 2\). Initially, through the binomial expansion, only the first two terms summed up to 2, showing equality. However, recognizing that additional positive terms exist confirms the inequality, tipping the balance to be greater than 2. This example emphasizes that inequalities help us determine bounds and relationships rather than just fixed values.

Algebra

Algebra is fundamental in understanding how numbers, variables, and operations fit together. In its essence, algebra provides rules and formulas to handle numbers even when they become more complex, such as in powers or expansions.
In the problem involving \((1.01)^{100}\), algebra plays a vital role in simplifying the expression using the Binomial Theorem. Here's how algebraic reasoning is applied:

• Understanding how to substitute values for \(a\) and \(b\) in the binomial formula.
• Calculating terms using algebraic manipulation and simplification.
• Using algebra to understand how additional positive terms affect the inequality.

These algebraic processes allow us to move from a straightforward expansion to proving an inequality, highlighting algebra’s power to simplify and elucidate complex expressions and relationships.