Problem 15 - Contemporary Precalculus

Question

Evaluate the expression. $$\left(\begin{array}{l}6 \\\0\end{array}\right)+\left(\begin{array}{l}6 \\\1\end{array}\right)+\left(\begin{array}{l}6\\\2\end{array}\right)+\left(\begin{array}{l}6 \\\3\end{array}\right)+\left(\begin{array}{l}6 \\\4 \end{array}\right)+\left(\begin{array}{l}6 \\\5\end{array}\right)+\left(\begin{array}{l}6 \\\6\end{array}\right)$$

Step-by-Step Solution

Answer

Answer: The sum of binomial coefficients for the power of 6 is 64.

Step 1: Rewrite the expression using binomial coefficients notation

The expression can be rewritten as: $$\binom{6}{0}+\binom{6}{1}+\binom{6}{2}+\binom{6}{3}+\binom{6}{4}+\binom{6}{5}+\binom{6}{6}$$

Step 2: Apply the binomial theorem with x=1 and y=1

Using the binomial theorem, we get: $$(1+1)^6 = \sum_{k=0}^6 \binom{6}{k}\cdot1^{6-k}\cdot1^k=\sum_{k=0}^6 \binom{6}{k}.$$

Step 3: Calculate the result

Now we can directly compute the expression: $$(1+1)^6 = 2^6 = 64$$ Therefore, the sum of these binomial coefficients is 64.

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool in algebra that allows us to expand expressions of the form \((x + y)^n\). It states that \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\). Here, \(\binom{n}{k}\) is the binomial coefficient, representing the number of ways to choose \(k\) items from \(n\) items.

In our original problem, we applied the binomial theorem using \(x=1\) and \(y=1\). This simplifies the expression to \((1+1)^6\), because when both \(x\) and \(y\) are 1, their powers will not affect the multiplication. The sum of the binomial coefficients \(\binom{6}{0} + \binom{6}{1} + ... + \binom{6}{6}\) simplifies to \(2^6\), which equals 64. Thus, it elegantly shows how the listing of all binomial coefficients corresponds to the expansion of a binary power.

Combinatorics

Combinatorics is the mathematics of counting and arranging objects. Binomial coefficients are key elements in combinatorial calculations. They tell us how many ways we can pick \(k\) items from a set of \(n\) items without considering the order.

In our exercise, each \(\binom{6}{k}\) is a combinatorial expression. For instance, \(\binom{6}{0}\) equals 1 because there is exactly one way to choose none out of six. Likewise, \(\binom{6}{3}\) computes how many ways you can pick 3 out of 6 items. It's defined by the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). Each term in the binomial sum represents a combination count that plays a vital role in combinatorics, illustrating how combinatorial ideas underpin many mathematical concepts.

Pascal's Triangle

Pascal's Triangle is a geometric representation of binomial coefficients, arranged in a triangular format. Each row 'n' represents the coefficients in the expansion of \((x+y)^n\). To find a binomial coefficient \(\binom{n}{k}\), locate the \(n\)-th row and the \(k\)-th position.

In the context of our exercise, the 7th row of Pascal's Triangle (since we start counting from 0) aligns with the expression \(\binom{6}{0} + \binom{6}{1} + ... + \binom{6}{6}\), representing 1, 6, 15, 20, 15, 6, 1. These numbers are the individual binomial coefficients that sum up as part of the Binomial Theorem application. Pascal's Triangle not only provides a simple way to find these coefficients but also showcases beautiful symmetry and patterns found in mathematics.