Problem 52 - Precalculus: Mathematical for Calculus

Question

Add each of the first five rows of Pascal's triangle, as indicated. Do you see a pattern? $$\begin{aligned} &\begin{array}{c} 1+1=? \\ 1+2+1=? \end{array}\\\ &1+3+3+1=?\\\ &1+4+6+4+1=?\\\ &1+5+10+10+5+1=? \end{aligned}$$Based on the pattern you have found, find the sum of the \(n\) th row: $$ \left(\begin{array}{l} n \\ 0 \end{array}\right)+\left(\begin{array}{l} n \\ 1 \end{array}\right)+\left(\begin{array}{l} n \\ 2 \end{array}\right)+\cdots+\left(\begin{array}{l} n \\ n \end{array}\right) $$ Prove your result by expanding \((1+1)^{n}\) using the Binomial Theorem.

Step-by-Step Solution

Answer

Each row's sum is a power of 2: the sum of the n-th row is \(2^n\).

Step 1: Calculate the Sum of Row 1

The first row of Pascal's Triangle is \(1+1\). Calculate the sum: \[1 + 1 = 2\]

Step 2: Calculate the Sum of Row 2

The second row of Pascal's Triangle is \(1+2+1\). Calculate the sum: \[1 + 2 + 1 = 4\]

Step 3: Calculate the Sum of Row 3

The third row of Pascal's Triangle is \(1+3+3+1\). Calculate the sum: \[1 + 3 + 3 + 1 = 8\]

Step 4: Calculate the Sum of Row 4

The fourth row of Pascal's Triangle is \(1+4+6+4+1\). Calculate the sum: \[1 + 4 + 6 + 4 + 1 = 16\]

Step 5: Calculate the Sum of Row 5

The fifth row of Pascal's Triangle is \(1+5+10+10+5+1\). Calculate the sum: \[1 + 5 + 10 + 10 + 5 + 1 = 32\]

Step 6: Observe the Pattern

The sums we calculated for the first five rows are: 2, 4, 8, 16, and 32. Observe that each sum is a power of 2: \(2^1, 2^2, 2^3, 2^4, 2^5\). This suggests that the sum of the elements of the \(n\)-th row is \(2^n\).

Step 7: Generalize for the n-th Row

The sum of the entries in the \(n\)-th row of Pascal's Triangle is:\[\left(\begin{array}{l} n \ 0 \end{array}\right)+\left(\begin{array}{l} n \ 1 \end{array}\right)+\cdots+\left(\begin{array}{l} n \ n \end{array}\right) = 2^n\]

Step 8: Prove Using Binomial Theorem

According to the Binomial Theorem, \((x+y)^n\) expands to:\[(x+y)^n = \left(\begin{array}{c} n \ 0 \end{array}\right)x^n + \left(\begin{array}{c} n \ 1 \end{array}\right)x^{n-1}y + \cdots + \left(\begin{array}{c} n \ n \end{array}\right)y^n\]By setting \(x = 1\) and \(y = 1\):\[(1+1)^n = \left(\begin{array}{c} n \ 0 \end{array}\right) + \left(\begin{array}{c} n \ 1 \end{array}\right) + \cdots + \left(\begin{array}{c} n \ n \end{array}\right)\]Which simplifies to:\[2^n = \left(\begin{array}{c} n \ 0 \end{array}\right) + \left(\begin{array}{c} n \ 1 \end{array}\right) + \cdots + \left(\begin{array}{c} n \ n \end{array}\right)\]

Key Concepts

Binomial Theorem

The Binomial Theorem is a mathematical formula that allows us to expand expressions of the form \((x+y)^n\). It's incredibly useful in Algebra and provides a way to systematically obtain each term. Here’s what it looks like:
\[(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\]
This means that for any non-negative integer \(n\), the expression \((x + y)^n\) can be expanded as the sum of \(n+1\) terms.

• Each term involves a coefficient obtained through combinations, noted as binomial coefficients \(\binom{n}{k}\).
• The coefficients are the entries of Pascal's Triangle.

By applying the Binomial Theorem to \((1+1)^n\), each coefficient corresponds to a number in Pascal's Triangle. Thus, we can also deduce that \((1+1)^n = 2^n\), where the sum of a row in Pascal’s Triangle results in \(2^n\). This offers a direct link between the Binomial Theorem and Pascal’s Triangle, revealing the sum pattern within the triangle.

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, combination, and permutation of sets of elements. In the context of Pascal's Triangle, it plays a crucial role in explaining how the elements are arranged.

• Each entry in Pascal’s Triangle can be expressed as a binomial coefficient.
• Binomial coefficients \(\binom{n}{k}\) tell us the number of ways to choose \(k\) elements from a set of \(n\) elements.

This is closely linked to the Binomial Theorem, where each coefficient corresponds to one term in the binomial expansion. For example:

• In the third row of Pascal's Triangle, \(1, 3, 3, 1\), the numbers come from \(\binom{3}{0}, \binom{3}{1}, \binom{3}{2}, \binom{3}{3}\).

So, combinatorics provides the foundation for understanding Pascal’s Triangle, summarizing how entries relate to combinations and how many subsets can be formed.

Pattern Recognition

Pattern recognition involves identifying regularities and relationships within a set of data. Romantic as it sounds, in mathematics, it helps us discover useful rules and shortcuts. In Pascal's Triangle, each row's sum follows a clear and predictable pattern.

• The first few sums were 2, 4, 8, 16, and 32.
• This pattern is consistent with the powers of 2: \(2^1, 2^2, 2^3, 2^4, 2^5\).
• Therefore, for any row, the sum of all the numbers is \(2^n\).

Recognizing this pattern not only helps in solving specific problems but also makes a larger statement about mathematical predictability. Through immediate observation, students can apply the power of 2 to quickly determine the sum of any row without direct computation!

Exponents

Exponents are a way to express repeated multiplication. They're central to understanding powers of numbers. For example, \(2^3\) is shorthand for \(2 \times 2 \times 2\) or 8.
Connecting exponents to the sums of Pascal’s Triangle underlines why the row sums align with powers of 2.

• When solving the sums for the first five rows of Pascal’s Triangle, each result is a power of 2.
• This directly reflects the expansion \((1+1)^n = 2^n\).
• Exponents provide a quick way to grasp this relation and makes calculations efficient.

Understanding exponents in this context allows us to make the leap from simply adding numbers to appreciating the inherent algebraic structure. As each row in Pascal’s Triangle adds one more power of 2, it represents exponential growth in a geometric sense. Thus, invoking the rules of exponents becomes an indispensable tool in recognizing and proving such patterns.