Problem 49 - Precalculus Mathematics for Calculus

Question

Simplify using the Binomial Theorem. Show that \((1.01)^{100}>2 . \quad\) [ 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 sum of the first two terms is 2, and additional positive terms make (1.01)^{100} > 2.

Step 1: Binomial Theorem Formula

Recall the Binomial Theorem which states that \((x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\). In our case, \(x = 1\), \(y = 0.01\), and \(n = 100\).

Step 2: First Term Calculation

Using the formula, the first term of the expansion is \(\binom{100}{0} \times 1^{100} \times (0.01)^0 = 1\).

Step 3: Second Term Calculation

The second term is \(\binom{100}{1} \times 1^{99} \times (0.01)^1 = 100 \times 0.01 = 1\).

Step 4: Sum of First Two Terms

Add the first two terms together: \(1 + 1 = 2\).

Step 5: Analyzing Increment Term

The third term of the expansion involves \(\binom{100}{2} \times 1^{98} \times (0.01)^2\), which gives a positive value because \( y^2 = (0.01)^2 > 0\). Thus, the entire sum must be slightly more than 2.

Step 6: Concluding Inequality

Thus, \((1.01)^{100}\) is greater than 2, since the sum of the first two terms already equals 2 and additional positive terms follow.

Key Concepts

Polynomial Expansion

The Binomial Theorem is a powerful tool for expanding expressions of the form \((x + y)^n\). It allows us to break down these expressions into a sum of terms, each involving the binomial coefficient \(\binom{n}{k}\). In simpler terms, the theorem helps us expand expressions into a series of terms based on their powers.
Each term in the expansion has the form \(\binom{n}{k} x^{n-k} y^k\). This means, for each term, you take:

• The binomial coefficient \(\binom{n}{k}\), which counts the ways to choose \(k\) items from \(n\).
• \(x\) raised to the power of \(n-k\).
• And \(y\) raised to the power of \(k\).

This is repeated for \(k\) from 0 to \(n\), producing all the terms in the series. For the exercise \((1.01)^{100} = (1 + 0.01)^{100}\), the first terms are easily calculated with small powers of 0.01, making the early terms the most significant.

Inequality Proof

Inequality proofs involve showing that one quantity is greater than or less than another. In our exercise, we aim to prove that \((1.01)^{100} > 2\). By starting with the Binomial Theorem expansion, we found that the sum of the first two significant terms is exactly 2.
The next step is to recognize the nature of the terms that follow. Each subsequent term in the expansion \(\binom{n}{k} \times (0.01)^k\) adds a positive number to the sum. The crux of the inequality lies in proving that more than just two terms contribute.Given that the third term \(\binom{100}{2} \times (0.01)^2\) and beyond remain positive, we conclude that the sum of all terms must be slightly more than 2, confirming the inequality.

Algebraic Simplification

Algebraic simplification entails reducing expressions to a simpler or more manageable form. It often involves combining like terms and simplifying coefficients. In this exercise, the task wasn't merely to plug numbers into a formula, but to take the sum of two calculated terms.
Starting with \(1 + 1 = 2\) derived from the Binomial expansion, we move to consider all subsequent terms. Each additional positive term in the series confirms that the simplified form is more than a mere sum of the first two terms. Keeping track of the coefficients and powers helps ensure that the focus remains clear on why the total is slightly above 2.By simplifying algebraic expressions effectively, complex concepts can be reduced to simple conclusions, supporting proofs and inequalities straightforwardly.