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Binomial Theorem: For any positive integer \(n\) and any real numbers \(x\) and \(y,\) \[(x + y)^n = \sum_{i=0}^n {n \choose i}x^{n-i}y^i\]

▼ Proof:
This is a proof by induction.

Base case: When \(n = 1,\) \((x+y)^1 = x + y.\) The sum is \begin{align} \sum_{i=0}^1 {1 \choose i}x^iy^{1-i} & = {1 \choose 0}x^{1-0}y^0 + {1 \choose 1}x^{1-1}y^1 \\ & = x + y \end{align} So, the theorem is true for \(n = 1.\)

When \(n = 2,\) \((x+y)^2 = x^2 + 2xy + y^2.\) The sum is \begin{align} \sum_{i=0}^2 {2 \choose i}x^iy^{2-i} & = {2 \choose 0}x^{2-0}y^0 + {2 \choose 1}x^{2-1}y^1 + {2 \choose 2}x^{2-2}y^2 \\ & = x^2 + 2xy + y^2 \end{align} So, the theorem is true for \(n = 2.\)

Inductive step: Suppose the theorem is true for \(n = k.\) We will show it is true for \(n = k+1.\) \begin{align} (x+y)^{k+1} & = (x+y)(x+y)^k \\ & = (x+y)\sum_{i=0}^k {k \choose i}x^{k-i}y^i \end{align} The last equality follows from the inductive hypothesis.

Next, distribute the sum over \(x+y.\) \begin{align} (x+y)\sum_{i=0}^k {k \choose i}x^{k-i}y^i & = \sum_{i=0}^k {k \choose i}x^{k-i+1}y^i + \sum_{i=0}^k {k \choose i}x^{k-i}y^{i+1} \\ & = \sum_{i=0}^k {k \choose i}x^{k-i+1}y^i + \sum_{i=1}^{k+1} {k \choose i-1}x^{k-i+1}y^i \\ & = x^{k + 1} + \sum_{i=1}^k {k \choose i}x^{k-i+1}y^i + \sum_{i=1}^k {k \choose i-1}x^{k-i+1}y^i + y^{k+1} \\ & = x^{k+1} + \sum_{i=1}^k \left[{k \choose i} + {k \choose i-1}\right]x^{k-i+1}y^i + y^{k+1} \\ \end{align} Next, compute the sum of combinations as follows: \begin{align} {k \choose i} + {k \choose i-1} & = \frac{k!}{i!(k-i)!} + \frac{k!}{(i-1)!(k-i+1)!} \\ & = \frac{(k-i+1)k!}{i!(k-i+1)!} + \frac{ik!}{(i!(k-i+1)!} \\ & = \frac{(k+1)k!}{i!(k-i+1)!} \\ & = \frac{(k+1)!}{i!(k+1-i)!} \\ & = {k+1 \choose i} \end{align} Using this identity, we have \begin{align} x^{k+1} + \sum_{i=1}^k \left[{k \choose i} + {k \choose i-1}\right]x^{k-i+1}y^i + y^{k+1} & = x^{k+1} + \sum_{i=1}^k {k+1 \choose i}x^{k+1-i}y^i + y^{k+1} \\ & = \sum_{i=0}^{k+1} {k+1 \choose i}x^{k-i+1}y^i \end{align} which completes the proof.

Application: Use the binomial theorem to expand \((2x+3)^4.\)

▼ Solution:
Use the binomial theorem with \(3x\) in place of \(x\) and \(3\) in place of \(y.\) \begin{align} (2x + 3)^4 & = {4 \choose 0}(2x)^4(3)^0 + {4 \choose 1}(2x)^3(3)^1 + {4 \choose 2}(2x)^2(3)^2 + {4 \choose 3}(2x)^1(3)^3 + {4 \choose 4}(2x)^0(3)^4 \\ & = {4 \choose 0}16x^4 + {4 \choose 1}24x^3 + {4 \choose 2}36x^2 + {4 \choose 3}18x + {4 \choose 4}81 \end{align} One can use Pascal's triangle to find the values of the combinations:
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

The combinations in row \(4\) are \(1, 4, 6, 4, 1.\) \begin{align} & {4 \choose 0}16x^4 + {4 \choose 1}24x^3 + {4 \choose 2}36x^2 + {4 \choose 3}18x + {4 \choose 4}81 \\ & = (1)16x^4 + (4)24x^3 + (6)36x^2 + (4)18x + (1)81 \\ & = 16x^4 + 96x^3 + 216x^2 + 72x + 81 \end{align}

Example: There are \(3\) bags. Each bag has \(2\) red balls and \(4\) yellow balls. A ball is drawn from each bag. Answer the following:

Solution: The number of ways to draw balls from bags can be represented with a polynomial. Each bag has \(2\) red balls and \(4\) yellow balls, so the polynomial representing a bag is \(2r + 4y\) where \(r\) is for red balls and \(y\) is for yellow balls.

There are \(3\) bags, so the polynomial that represents one ball from each bag is \[(2r+4y)^3.\] By the binomial theorem, \begin{align} (2r+4y)^3 & = {3 \choose 0}(2r)^3(4y)^0 + {3 \choose 1}(2r)^2(4y)^1 + {3 \choose 2}(2r)^1(4y)^2 + {3 \choose 3}(2r)^0(4y)^3 \\ & = (1)(8r^3)(1) + (3)(4r^2)(4y) + (3)(2r)(16y^2) + (1)(1)(64y^3) \\ & = 8r^3 + 48r^2y + 96ry^2 + 64y^3 \end{align} The coefficients tell the number of outcomes for each possible outcome. For example, \(8r^3\) indicates there are \(8\) ways to draw \(3\) red balls. The answers to the questions are

Fill in the coefficients to \((2x-7)^4.\)

 

What is the coefficient of \(x^7y^3\) in \((-3x-4y)^{10}?\)