Problem 105 - Essential Calculus

Question

Let \(x>0\) and \(n>1\) be real numbers. Prove that \((1+x)^{n}>1+n x\).

Step-by-Step Solution

Answer

By using the Binomial theorem, we have established that \((1+x)^{n}>1+n x\), given that \(x > 0\) and \(n > 1\), proving the given inequality is true.

Step 1: Understanding the binomial theorem

The Binomial Theorem tells us how to expand expressions of the form \((a+b)^{n}\) for any integers \(n\). According to it, we have \((1+x)^{n} = \sum_{k=0}^{n} {n\choose{k}} x^{k}\), where \({n\choose{k}}\) are binomial coefficients. When \(k=0\), this term will reduce to 1, when \(k=1\), it results in \(n\).

Step 2: Apply the binomial theorem

By applying the binomial theorem, we can expand the left side of the given expression as \((1+x)^{n} = 1+ nx + \sum_{k=2}^{n} {n\choose{k}} x^{k}\). Since, each term in the series \(\sum_{k=2}^{n} {n\choose{k}} x^{k}\) is positive as \(x > 0\) and \(n > 1\), we can say that \((1+x)^{n} > 1+ nx\).

Step 3: Finalize the proof

Since each term in the series \(\sum_{k=2}^{n} {n\choose{k}} x^{k}\) is positive, it must be the case that \((1+x)^{n}\) is greater than \(1+ nx\), exactly what we wanted to prove.

Key Concepts

Binomial Coefficients

Binomial coefficients are central to expressions expanded by the Binomial Theorem. They appear in the form \( {n \choose k} \), and are used to calculate the coefficients of terms in the expansion of \((a+b)^n\). These coefficients are determined using the formula \({n\choose k} = \frac{n!}{k!(n-k)!}\), where \(!\) represents factorial, indicating the product of all positive integers up to that number.

• When \(k=0\), the coefficient is always \(1\), representing the term \(a^n\).
• For \(k=1\), the coefficient becomes \(n\), and the term represents an expression like \(na^{n-1}b\).
• Each subsequent coefficient involves more complex interactions between \(a\) and \(b\), increasing in complexity as \(k\) increases.

In the context of our problem, understanding binomial coefficients allows us to recognize that as \(x > 0\) and \(n > 1\), additional positive terms are introduced when expanding \((1+x)^n\), thus making it greater than \(1 + nx\).

Inequality Proofs

Inequality proofs involve demonstrating that one mathematical expression is greater (or less) than another. These proofs often involve algebraic manipulation, logical reasoning, or applying known theorems.

• One common technique is direct comparison, where expressions are rearranged or expanded to directly compare terms.
• Another method is indirect proof, such as proof by contradiction, where you assume the opposite of what you wish to prove and show this leads to an inconsistency.

In this exercise, applying the binomial theorem allowed us to expand \((1+x)^n\) in such a way that it obviously becomes larger than \(1 + nx\). All the extra terms, which arise for \(k \geq 2\), are positive due to \(x > 0\). This essentially adds to the inequality proving \((1+x)^n > 1 + nx\).

Exponential Expressions

Exponential expressions involve expressions where a constant base is raised to a variable exponent, like \( (1+x)^n \). These types of expressions are fundamental in mathematics, often simplifying or representing complex real-world phenomena succinctly.

• An exponential expression can show how a quantity grows over time or under certain conditions.
• In inequalities, exponential terms may greatly influence the comparison, especially as the exponent increases.

For the proof, the exponential expression \((1+x)^n\) is critical to understanding growth. As \(n\) increases, the expression generally captures faster growth compared to linear terms like \(1 + nx\).
Utilizing properties of exponentials and binomial expansions, you can demonstrate how additional terms make \((1+x)^n\) surpass \(1+nx\). This reflects the significant impact of exponential growth, even when the base \(x\) is small but positive.