Problem 41 - Discrete Mathematics with Applications

Question

From the binomial expansion \((1+x)^{n}=\sum_{r=0}^{n}\left(\begin{array}{l}{n} \\\ {r}\end{array}\right) x^{r},\) it can be shown using calculus that \(n(1+x)^{n-1}=\sum_{r=1}^{n}\left(\begin{array}{c}{n} \\\ {r}\end{array}\right) r x^{n-1}\) . Using this result, prove each. $$ 1\left(\begin{array}{l}{n} \\\ {1}\end{array}\right)+2\left(\begin{array}{l}{n} \\\ {2}\end{array}\right)+3\left(\begin{array}{l}{n} \\\ {3}\end{array}\right)+\cdots+n\left(\begin{array}{l}{n} \\\ {n}\end{array}\right)=n 2^{n-1} $$

Step-by-Step Solution

Answer

By plugging in \(x=1\) in the given calculus-based expansion and simplifying the expression, we get \(n(2)^{n-1}=\sum_{r=1}^{n}\binom{n}{r} r\). This proves the given expression: $$ 1\left(\begin{array}{l}{n} \\\ {1}\end{array}\right)+2\left(\begin{array}{l}{n} \\\ {2}\end{array}\right)+3\left(\begin{array}{l}{n} \\\ {3}\end{array}\right)+\cdots+n\left(\begin{array}{l}{n} \\\ {n}\end{array}\right)=n 2^{n-1} $$

Step 1: Write down the given expansions

We have the following expansions: 1. Binomial expansion: \((1+x)^{n}=\sum_{r=0}^{n}\binom{n}{r} x^{r}\) 2. Calculus-based expansion: \(n(1+x)^{n-1}=\sum_{r=1}^{n}\binom{n}{r} r x^{r-1}\)

Step 2: Plug in the value x = 1 in the calculus-based expansion

Replace \(x\) with \(1\) in the calculus-based expansion: \(n(1+1)^{n-1}=\sum_{r=1}^{n}\binom{n}{r} r\)

Step 3: Simplify

Simplify the expression: \(n(2)^{n-1}=\sum_{r=1}^{n}\binom{n}{r} r\)

Step 4: Interpret the result

We have obtained the needed result which proves the given expression: $$ 1\left(\begin{array}{l}{n} \\\ {1}\end{array}\right)+2\left(\begin{array}{l}{n} \\\ {2}\end{array}\right)+3\left(\begin{array}{l}{n} \\\ {3}\end{array}\right)+\cdots+n\left(\begin{array}{l}{n} \\\ {n}\end{array}\right)=n 2^{n-1} $$ The result is proven by plugging in \(x=1\) in the given calculus-based expansion and simplifying the expression.

Key Concepts

Binomial Coefficients

Binomial coefficients are an essential part of the binomial theorem, which allows for the expansion of expressions of the form \((1 + x)^n\). These coefficients, typically denoted as \(\binom{n}{r}\), represent the number of ways to choose \(r\) items out of \(n\) without regard to order.

They are calculated using the formula:

• \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\)

Where \(!\) denotes factorial, meaning the product of an integer and all the integers below it. Binomial coefficients appear throughout combinatorics, probability, and algebra, making them a cornerstone in many mathematical computations.

In the context of the binomial expansion, they are used to determine the coefficients of each term in the expansion of \((1+x)^n\):

• First term: \(\binom{n}{0}x^0\)
• Second term: \(\binom{n}{1}x^1\)
• ...until...
• Last term: \(\binom{n}{n}x^n\)

This organization of terms allows for systematic calculation and discovery of important properties within a binomial distribution.

Calculus-Based Expansion

Using calculus, the binomial expansion can be differentiated to yield significant insights into its structure. Starting with the binomial expansion formula \((1+x)^n\), if we differentiate both sides with respect to \(x\), we obtain:

• \(n(1+x)^{n-1}\)

This result leads us to a calculus-based expansion equation:

• \(n(1+x)^{n-1} = \sum_{r=1}^{n}\binom{n}{r} r x^{r-1}\)

Each term on the right-hand side arises from differentiating the corresponding term in the original binomial expression. When \(x = 1\) is substituted into this derivative-based expansion, the sum becomes a simple calculation:

• \(n(2)^{n-1} = \sum_{r=1}^{n}\binom{n}{r}r\)

This operation transforms the expression and allows us to equate it to prove powerful identities involving sums of weighted binomial coefficients, as seen in verifying the given identity \(1\binom{n}{1} + 2\binom{n}{2} + \cdots + n\binom{n}{n} = n 2^{n-1}\).

Combinatorial Proofs

Combinatorial proofs provide an alternate way to understand or demonstrate a mathematical claim using the principles of combinatorics. Rather than relying on algebraic manipulation or calculus, these proofs relate the problem to counting arguments or combinations.

The identity \(1\binom{n}{1} + 2\binom{n}{2} + \cdots + n\binom{n}{n} = n 2^{n-1}\) can be approached with such combinatorial insights.

In this context, each term \(r\binom{n}{r}\) can be interpreted as the number of ways to select a group of \(r\) elements and designate a specific one as a leader or a special role within the group.

• Think of \(r\binom{n}{r}\) as choosing \(r\) leaders from \(n\) options.

Such interpretations often stem from visualizing combinations or permutations in a tangible setup.

These proofs offer a conceptual clarity that complements algebraic methods, revealing connections between different fields in math and solidifying understanding through everyday counting techniques. This intuitive reasoning makes seemingly complex formulas accessible, reinforcing why a mathematical statement is fundamentally true.