Question

Apply calculus and the binomial theorem to \((1+x)^{n}\) to show that $$ \left(\begin{array}{l} n \\ 1 \end{array}\right)+2\left(\begin{array}{l} n \\ 2 \end{array}\right)+3\left(\begin{array}{l} n \\ 3 \end{array}\right)+\cdots=n 2^{n-1} . $$

Short answer

Using calculus and the binomial theorem, we show that \(1\cdot\binom{n}{1} + 2\cdot\binom{n}{2} + 3\cdot\binom{n}{3} + \ldots = n 2^{n-1}\).

Step-by-step solution

Understand the Binomial Theorem

The binomial theorem states that for any non-negative integer \(n\) and any real number \(x\), \((1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k\). Each term \(\binom{n}{k}\) represents the number of ways to choose \(k\) elements from a set of \(n\) elements.

Differentiate the Expansion

Differentiate \((1+x)^n\) with respect to \(x\). Since \((1+x)^n\) is now \(nx(1+x)^{n-1}\) upon differentiating and applying the chain rule.

Apply the Binomial Theorem to the Derivative

The derivative from Step 2 gives us \(n(1+x)^{n-1} = \sum_{k=1}^n \binom{n}{k} k x^{k-1}\). Here, the differentiation reduces the exponent of \(x\) by 1, and the power rule brings down the factor \(k\).

Set x = 1 in the Expression

Substitute \(x = 1\) into the differentiated binomial expansion, \(n(1+1)^{n-1} = \sum_{k=1}^n \binom{n}{k} k \). Thus, this simplifies to \(n 2^{n-1} = \sum_{k=1}^n k \binom{n}{k}\).

Identify the Desired Result

Recognize that \( \sum_{k=1}^n k \binom{n}{k} \) matches the left-hand side of our original equation: \(1\cdot\binom{n}{1} + 2\cdot\binom{n}{2} + 3\cdot\binom{n}{3} + \ldots = n 2^{n-1}\). This confirms the equation.

Key concepts

Calculus

Calculus is a branch of mathematics that provides tools to study changes. It helps in understanding how things evolve, whether it's in physical space, financial markets, or populations. One of the foundational concepts in calculus is the notion of a limit, which leads to the idea of derivatives and integrals. For this exercise, calculus is applied to understand how functions change as their inputs vary. The application of calculus assists in analyzing the binomial expansion and differentiating it step by step to prove complex equations. Thus, it plays a crucial role in providing a deeper insight into problems involving changing quantities.

Differentiation

Differentiation is a core concept in calculus that deals with finding the instantaneous rate of change of a function with respect to one of its variables. In simple terms, differentiation gives us the slope of the tangent line to the function's graph at any given point. Here, it is applied to the binomial expansion

1. First, we recognize the pattern in which the function changes.
2. Then, by differentiating, we shift focus to how each term in the expansion expands or shrinks as the variable changes.
3. In this exercise, differentiation shows how an initial power function, like \( (1+x)^{n} \), behaves when its input changes, revealing a new relationship.

This operation is key to arriving at the simplified expression that matches the required result.

Combinatorics

Combinatorics is the branch of mathematics dealing with the study of finite or countable discrete structures. It encompasses various techniques and theorems used to count, organize, and analyze combinations, permutations, and other structures without exhausting all possibilities. In this exercise, the binomial coefficients \(\binom{n}{k}\) from the binomial theorem are used:

• Each coefficient \(\binom{n}{k}\) represents the number of ways to choose \k\ components from \ components.
• This concept provides the foundation for analyzing terms in the binomial expansion.
• Understanding combinatorics helps make sense of mathematical proofs involving combinations such as those found in the binomial expansion.

With these principles, one can grasp how to predict and analyze the patterns and coefficients emerging from the binomial expansion.

Mathematical Proof

A mathematical proof is a logical argument that verifies the truth of a mathematical statement. Proofs are essential because they provide a solid foundation upon which the structure of mathematics is built. In this exercise:

• We begin with the hypothesis—the binomial theorem and its subsequent differentiation.
• Steps in the proof demonstrate each transition clearly from concept to differentiated expression to simplified end forms.
• Substituting \(x = 1\) is a crucial step that wraps up the argument logically.

The goal is to connect and verify that \(n 2^{n-1}\) matches the numeric transformation given by the original summation. This proof showcases a blend of algebra, calculus, and combinatorics, illustrating how various mathematical disciplines interconnect.