Problem 58 - Algebra and Trigonometry

Question

Sums of Binomial Coefficients Add each of the first five rows of Pascal’s triangle, as indicated. Do you see a pattern? $$ \begin{array}{c}{1+1=?} \\ {1+2+1=?} \\ {1+3+3+1=?} \\ {1+4+6+4+1=?} \\\ {1+5+10+10+5+1=?}\end{array} $$ On the basis of the pattern you have found, find the sum of the nth 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

The sum of the nth row is \(2^n\).

Step 1: Identify the Rows of Pascal's Triangle

Pascal's Triangle starts with the top row as row 0: \(\{1\}\), row 1: \(\{1, 1\}\), row 2: \(\{1, 2, 1\}\), row 3: \(\{1, 3, 3, 1\}\), row 4: \(\{1, 4, 6, 4, 1\}\), and row 5: \(\{1, 5, 10, 10, 5, 1\}\). We will add the elements of these rows.

Step 2: Add Elements in Each Row

Add all the numbers in each row.- Row 0: \(1 = 1\)- Row 1: \(1 + 1 = 2\)- Row 2: \(1 + 2 + 1 = 4\)- Row 3: \(1 + 3 + 3 + 1 = 8\)- Row 4: \(1 + 4 + 6 + 4 + 1 = 16\)- Row 5: \(1 + 5 + 10 + 10 + 5 + 1 = 32\)

Step 3: Observe the Pattern

The sums observed are: 1, 2, 4, 8, 16, 32. These numbers are powers of 2: \(2^0, 2^1, 2^2, 2^3, 2^4, 2^5\). We observe that the sum of numbers in the nth row is \(2^n\).

Step 4: Express the General Pattern

For the nth row in Pascal's Triangle, the sum of binomial coefficients is \(2^n\). This can be written as:\[\sum_{k=0}^{n} \binom{n}{k} = 2^n\]

Step 5: Prove the Pattern Using the Binomial Theorem

The Binomial Theorem states that \[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k\]By setting \(a = 1\) and \(b = 1\),\[(1+1)^n = \sum_{k=0}^{n} \binom{n}{k}1^{n-k}1^k = 2^n\]Hence, we have proved the sum of the nth row's binomial coefficients equals \(2^n\).

Key Concepts

Binomial Coefficients

Binomial coefficients are crucial elements found in mathematics, particularly in algebra and combinatorics. You might have seen them as numbers in Pascal's Triangle, like the first few rows: \( \{1\} \), \( \{1, 1\} \), \( \{1, 2, 1\} \), and so on. They represent the number of ways to choose a subset of elements from a larger set, more formally represented as \( \binom{n}{k} \), which reads as "n choose k." This formula is defined as:

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

where \( n! \) is the factorial of \( n \). The factorial of a number is the product of all positive integers up to that number.
For example, \( \binom{4}{2} \) would be calculated as \( \frac{4!}{2!2!} = 6 \), meaning there are 6 ways to choose 2 elements from a set of 4.
Understanding binomial coefficients helps us not just in defining combinations but also in exploring patterns in Pascal’s Triangle, as we'll see further on.

Binomial Theorem

The Binomial Theorem provides a powerful way to expand expressions that are raised to a power. In particular, it’s useful for expanding expressions like \((a+b)^n\). The theorem states:

• \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

This means that when you expand \((a+b)^n\), each term in the expansion will involve binomial coefficients, powers of \(a\), and powers of \(b\).
For instance, applying the theorem to expand \((x+y)^2\) gives us:

• \( \binom{2}{0}x^2y^0 + \binom{2}{1}x^1y^1 + \binom{2}{2}x^0y^2 = 1\cdot x^2 + 2\cdot xy + 1\cdot y^2 = x^2 + 2xy + y^2 \)

In terms of the problem, by setting \(a = 1\) and \(b = 1\) in the binomial theorem, we showed how the sum of the binomial coefficients of the nth row is equal to \(2^n\). This fascinating result shows how these simple binomial coefficients are deeply connected to exponential growth sequences.

Mathematical Patterns

Mathematics is all about finding and understanding patterns. In the exercise, adding the numbers in each row of Pascal's Triangle reveals a fascinating pattern. The sums form the sequence: 1, 2, 4, 8, 16, 32. At a glance, these numbers might seem familiar—they are powers of 2.
This observation is not just a random coincidence. Each row of Pascal's Triangle corresponds to a particular power of 2, specifically \(2^n\) for the nth row. When you add elements of a row, you essentially sum up all combinations of the set, which, as proven through the binomial theorem, results in calculating \((1+1)^n = 2^n\).

• Row 0: \(1 = 2^0\)
• Row 1: \(1 + 1 = 2^1\)
• Row 2: \(1 + 2 + 1 = 2^2\)
• And so on...

Recognizing these mathematical patterns helps simplify complex problems and reveals deeper insights into the structure and behavior of numbers. Whether dealing with algebra, geometry, or calculus, pattern recognition is often the first step towards a solution.