Problem 52 - 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? $$ \begin{array}{c}{1+1=?} \\ {1+2+1=?} \\ {1+3+3+1=?} \\ {1+4+6+4+1=?} \\\ {1+5+10+10+5+1=?}\end{array} $$ Based on 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: Examine and Sum Each Row

Let's find the sum of numbers in each specified row of Pascal's Triangle.\1st row: \( 1 + 1 = 2 \)\2nd row: \( 1 + 2 + 1 = 4 \)\3rd row: \( 1 + 3 + 3 + 1 = 8 \)\4th row: \( 1 + 4 + 6 + 4 + 1 = 16 \)\5th row: \( 1 + 5 + 10 + 10 + 5 + 1 = 32 \)

Step 2: Identify the Pattern

By examining the sums calculated in Step 1, we observe a pattern: 2, 4, 8, 16, 32. These numbers suggest powers of 2: \(2^1, 2^2, 2^3, 2^4, 2^5\). Thus, the sum of the \(n\)th row is \(2^n\).

Step 3: Generalize the Pattern for the nth Row

From the identified pattern, for the \(n\)th row, the sum of binomial coefficients is given by \(2^n\). Therefore, \(\sum_{k=0}^n \binom{n}{k} = 2^n\)\.

Step 4: Prove Using Binomial Theorem

According to the Binomial Theorem, \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\).\ Set \(x = 1\) and \(y = 1\), then \((1 + 1)^n = \sum_{k=0}^n \binom{n}{k} 1^{n-k} 1^k = \sum_{k=0}^n \binom{n}{k}\).\ Hence, we get \(2^n = \sum_{k=0}^n \binom{n}{k}\).

Key Concepts

Pascal's Triangle

Pascal's Triangle is a fascinating mathematical concept where every row represents the coefficients of the binomial expansion. To construct it, start with a "1" at the top and each new row begins and ends with "1". Every element inside the triangle is the sum of the two directly above it from the row before.

• The 0th row is just 1.
• The 1st row is 1, 1.
• The 2nd row is 1, 2, 1.
• This pattern continues indefinitely.

Pascal's Triangle is not only a nice pattern but also a practical tool in various computations. What's truly amazing about it, especially in this exercise, is that the sum of the numbers in each row equals a power of 2. For instance, the sum of the 3rd row is 8, which is \(2^3\). This insight becomes pivotal when connecting these sums to the Binomial Theorem.

Binomial Theorem

The Binomial Theorem provides a formula for expanding any power of a binomial. A binomial is simply a mathematical expression with two terms like \((x + y)\). When raised to a power \(n\), its expansion can be found using the Binomial Theorem:
\[ (x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k \]

• \(\binom{n}{k}\) is the binomial coefficient, which is just the number found at position \(k\) in row \(n\) of Pascal's Triangle.
• The exponents \(n-k\) and \(k\) show how many times you take the first term \(x\) and the second term \(y\) respectively.

For instance, the coefficient \(\binom{4}{2}\) from the 4th row and position 2, is used to expand \((x + y)^4\). This is why each row of Pascal's Triangle not only lists these coefficients but adds up to a power of 2 when you set both \(x\) and \(y\) as 1 - leading to \((1+1)^n = 2^n\).

Powers of 2

Powers of 2, such as 2, 4, 8, 16, and so on, are numbers obtained by multiplying 2 by itself a certain number of times. They are highly significant in mathematics and its applications because of their simple binary representation.

• \(2^1 = 2\)
• \(2^2 = 4\)
• \(2^3 = 8\)
• and so on...

In the exercise, discovering that the sum of the binomial coefficients row in Pascal's Triangle matches these powers of 2 deepens our understanding. This connection directly links binomial expansion with these sums, demonstrating the neat symmetry and beauty of mathematical structures. When you calculate the sum of the coefficients by expansion \((x + y)^n\), and by substitution, choosing \(x = 1\) and \(y = 1\), it simplifies to \((1+1)^n = 2^n\). That's why each row's sum in Pascal's Triangle is directly related to the powers of 2.