1.6 The Binomial Theorem $(a+b)^n$

1. The Symbol $n!$ and $nCr$

The Factorial $n!$

A symbol called "n factorial" is defined by $n! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot n$.

  • $1! = 1$
  • $2! = 1 \cdot 2 = 2$
  • $3! = 1 \cdot 2 \cdot 3 = 6$
  • $4! = 1 \cdot 2 \cdot 3 \cdot 4 = 24$
  • $5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120$
  • Definition: $0! = 1$

The Symbol $nCr$ or $\binom{n}{r}$

This symbol is read "n choose r"; it is given by the formula:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

For example:

  • "5 choose 2" or $5C2$: $\binom{5}{2} = \frac{5!}{2!3!} = 10$
  • "10 choose 3" or $10C3$: $\binom{10}{3} = \frac{10!}{3!7!} = 120$

Notice: $\binom{5}{2}$ represents the number of ways to choose 2 items out of 5 (e.g., choosing 2 letters from A, B, C, D, E yields 10 combinations).

⚠️ GDC Usage

The GDC can be used for these calculations:

  • Factorial: MENU $\rightarrow$ OPTN $\rightarrow$ PROB $\rightarrow$ $x!$
  • Combinations: MENU $\rightarrow$ OPTN $\rightarrow$ PROB $\rightarrow$ $nCr$. (e.g., for $5C2$, press 5, then nCr, then 2).

2. The Binomial Theorem (Practically)

The coefficients of the expansion of $(a+b)^n$ may be obtained by Pascal's Triangle:

1
1   1
1   2   1   $\rightarrow$ coefficients of $(a+b)^2$
1   3   3   1   $\rightarrow$ coefficients of $(a+b)^3$
1   4   6   4   1   $\rightarrow$ coefficients of $(a+b)^4$
1   5   10   10   5   1   $\rightarrow$ coefficients of $(a+b)^5$

To write the formula for $(a+b)^5$:

  1. Step 1: Write down the variable terms. Exponents of $a$ decrease from 5 to 0, while exponents of $b$ increase from 0 to 5. (Sum of exponents is always 5).
    $a^5b^0, \quad a^4b^1, \quad a^3b^2, \quad a^2b^3, \quad a^1b^4, \quad a^0b^5$
  2. Step 2: Obtain the coefficients from Pascal's triangle (1, 5, 10, 10, 5, 1).

Therefore:

$(a+b)^5 = 1a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + 1b^5$

3. The Binomial Theorem (Formally)

Another way to obtain the coefficients is by using the symbol $\binom{n}{r}$. In general:

$(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \dots + \binom{n}{n}a^0 b^n$

The general term is given by: $\binom{n}{r} a^{n-r} b^r$

EXAMPLE 1

Find the expansions of $(2x+3)^3$ and $(2x-3)^3$.

Solution:

Apply the theorem for $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:
$(2x+3)^3 = (2x)^3 + 3(2x)^2(3) + 3(2x)(3^2) + 3^3$
$= 8x^3 + 36x^2 + 54x + 27$
For $(2x-3)^3$, use substitution $a=2x$ and $b=-3$ (or use alternating signs):
$(2x-3)^3 = 8x^3 - 36x^2 + 54x - 27$

EXAMPLE 2

Expand $(2x-3)^4$.

Solution:

Using the formula for $(a-b)^4$ (coefficients 1, 4, 6, 4, 1 and alternating signs):
$(2x-3)^4 = (2x)^4 - 4(2x)^3(3) + 6(2x)^2(3)^2 - 4(2x)(3)^3 + 3^4$
$= 16x^4 - 96x^3 + 216x^2 - 216x + 81$

4. Finding Particular Terms

Questions often ask for a specific term rather than the whole expansion.

EXAMPLE 3

In the expansion of $(2x-3)^4$ find the term of $x^3$.

Solution:

Follow three steps:
Step 1: Write down the two terms: $(2x)$ and $(-3)$.
Step 2: Split the exponent $n=4$ to get $x^3$: $(2x)^3(-3)^1$.
Step 3: Attach the coefficient $\binom{4}{1}$ (where $r=1$ is the exponent of $b$).
Result: $\binom{4}{1}(2x)^3(-3)^1 = 4(8x^3)(-3) = \mathbf{-96x^3}$.

EXAMPLE 4

Find the term of $x^5$ in the expansion of $(2x-3)^7$.

Solution:

The term is $\binom{7}{2}(2x)^5(-3)^2$.
Calculation: $21(32x^5)(9) = \mathbf{6048x^5}$.

EXAMPLE 5

In the expansion of $(2x^2+1)^8$ find the coefficient of $x^{10}$.

Solution:

Split 8 appropriately to obtain $x^{10}$ from $(2x^2)$: Exponent must be 5 (since $(x^2)^5 = x^{10}$).
Term: $\binom{8}{3}(2x^2)^5(1)^3$
Calculation: $56(32)x^{10}(1) = \mathbf{1792x^{10}}$.

EXAMPLE 6

In the expansion of $(2x+\frac{1}{x})^6$ find (a) the coefficient of $x^2$ and (b) the constant term.

Solution:

(a) The term is $\binom{6}{2}(2x)^4(\frac{1}{x})^2 = 15(16x^4)(\frac{1}{x^2}) = 240x^2$.
Coefficient is 240.
(b) The constant term is the coefficient of $x^0$. To eliminate $x$, split $n=6$ into 3 and 3.
$\binom{6}{3}(2x)^3(\frac{1}{x})^3 = 20(8x^3)(\frac{1}{x^3}) = \mathbf{160}$.

EXAMPLE 7

Find the constant term in the expansion of $(2x^2 - \frac{3}{x})^{12}$.

Solution:

Terms are $(2x^2)$ and $(-\frac{3}{x})$.
To eliminate $x$, use exponents 4 and 8: $(x^2)^4 = x^8$ and $(\frac{1}{x})^8 = \frac{1}{x^8}$.
Term: $\binom{12}{8}(2x^2)^4(-\frac{3}{x})^8 = 495(16)(6561) = \mathbf{51963120}$.

EXAMPLE 8

Find the term of $x^5$ in the expansion of $(2x+3)(4x+1)^7$.

Solution:

In the expansion of $(4x+1)^7$, two terms are needed:
1. The term of $x^4$ (to be multiplied by $2x$):
$\binom{7}{3}(4x)^4(1)^3 = 35(256)x^4 = 8960x^4$.
2. The term of $x^5$ (to be multiplied by $3$):
$\binom{7}{2}(4x)^5(1)^2 = 21(1024)x^5 = 21504x^5$.
Final term: $2x(8960x^4) + 3(21504x^5) = 17920x^5 + 64512x^5 = \mathbf{82432x^5}$.

EXAMPLE 9 (Mainly for HL)

(a) Verify that $2x^2-3x-2 = (2x+1)(x-2)$.
(b) Find the coefficient of $x^2$ in the expansion of $(2x^2-3x-2)^5$.

Solution:

(a) $(2x+1)(x-2) = 2x^2 - 4x + x - 2 = 2x^2 - 3x - 2$.
(b) Find the coefficient of $x^2$ in $(2x+1)^5(x-2)^5$. Expand each factor up to $x^2$:
$(1+2x)^5 = 1 + 10x + 40x^2 + \dots$
$(-2+x)^5 = (-2)^5 + 5(-2)^4x + \binom{5}{2}(-2)^3x^2 + \dots = -32 + 80x - 80x^2 + \dots$
Multiply the expansions: $(1+10x+40x^2)(-32+80x-80x^2)$.
Identify $x^2$ terms: $1(-80x^2) + (10x)(80x) + (40x^2)(-32)$.
$-80 + 800 - 1280 = \mathbf{-560}$.