1.6 Binomial Theorem

1. The Symbol $n!$ and $^nC_r$

The Factorial $n!$

A symbol called "n factorial" is defined by the product of all positive integers less than or equal to $n$:

$$n! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot n$$

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

The Symbol $^nC_r$ or $\binom{n}{r}$

This symbol is read as "n choose r" and is given by the formula:

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

For example:

  • "5 choose 2" or $^5C_2$: $\displaystyle \binom{5}{2} = \dfrac{5!}{2!3!} = \dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{(2 \cdot 1)(3 \cdot 2 \cdot 1)} = 10$
  • "10 choose 3" or $^{10}C_3$: $\displaystyle \binom{10}{3} = \dfrac{10!}{3!7!} = \dfrac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120$

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

2. The Binomial Theorem (Practically)

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

1 $(a+b)^0$
11 $(a+b)^1$
121 $(a+b)^2$
1331 $(a+b)^3$
14641 $(a+b)^4$
15101051 $(a+b)^5$

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

  • Step 1: Write down the variable terms. The exponents of $a$ decrease from $5$ to $0$, while the exponents of $b$ increase from $0$ to $5$. (Notice the sum of the exponents is always $5$).

    $$a^5b^0, \quad a^4b^1, \quad a^3b^2, \quad a^2b^3, \quad a^1b^4, \quad a^0b^5$$

  • Step 2: Obtain the coefficients from the corresponding row of Pascal's triangle ($1, 5, 10, 10, 5, 1$) and attach them to the terms.

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 without writing out Pascal's triangle is by using the combination 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 of the expansion is given by:

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

Example 1

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

Solution:

Apply the binomial theorem structure for $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$$\begin{aligned} (2x+3)^3 &= (2x)^3 + 3(2x)^2(3) + 3(2x)(3^2) + 3^3 \\ &= 8x^3 + 36x^2 + 54x + 27 \end{aligned}$$

For $(2x-3)^3$, use the substitution $a=2x$ and $b=-3$. This simply causes the signs to alternate:

$$\begin{aligned} (2x-3)^3 &= (2x)^3 + 3(2x)^2(-3) + 3(2x)(-3)^2 + (-3)^3 \\ &= 8x^3 - 36x^2 + 54x - 27 \end{aligned}$$

Example 2

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

Solution:

Using the formula for $(a-b)^4$ (coefficients $1, 4, 6, 4, 1$ and alternating signs):

$$\begin{aligned} (2x-3)^4 &= (2x)^4(-3)^0 + 4(2x)^3(-3)^1 + 6(2x)^2(-3)^2 + 4(2x)^1(-3)^3 \\ &+ (2x)^0(-3)^4 \\ &= 16x^4 - 4(8x^3)(3) + 6(4x^2)(9) - 4(2x)(27) + 81 \\ &= 16x^4 - 96x^3 + 216x^2 - 216x + 81 \end{aligned}$$

4. Finding Particular Terms (The General Term Method)

Questions in IB exams often ask for a specific term or coefficient rather than the whole expansion. The safest and most algebraic way to find it is to use the general term formula: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.

Example 3

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

Solution:

Write out the general term for $n=4$, $a=2x$, and $b=-3$:

$$T_{r+1} = \binom{4}{r} (2x)^{4-r} (-3)^r$$

To obtain the $x^3$ term, we must set the power of $x$ equal to $3$:

$$4 - r = 3 \implies r = 1$$

Substitute $r=1$ back into the general term:

$$ T_2 = \binom{4}{1} (2x)^3 (-3)^1 = 4 \cdot (8x^3) \cdot (-3) = -96x^3 $$

Example 4

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

Solution:

Write out the general term:

$$T_{r+1} = \binom{7}{r} (2x)^{7-r} (-3)^r$$

We need the power of $x$ to be $5$:

$$7 - r = 5 \implies r = 2$$

Substitute $r=2$:

$$ T_3 = \binom{7}{2} (2x)^5 (-3)^2 = 21 \cdot (32x^5) \cdot 9 = 6048x^5 $$

Example 5

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

Solution:

Write out the general term and group the exponents of $x$:

$$ T_{r+1} = \binom{8}{r} (2x^2)^{8-r} (1)^r = \binom{8}{r} 2^{8-r} \cdot x^{2(8-r)} \cdot 1 $$

Equate the exponent of $x$ to $10$:

$$16 - 2r = 10 \implies 2r = 6 \implies r = 3$$

Substitute $r=3$ to find the term:

$$ T_4 = \binom{8}{3} 2^{8-3} \cdot x^{10} = 56 \cdot 32 \cdot x^{10} = 1792x^{10} $$

The coefficient is 1792.

Example 6

In the expansion of $\left(2x+\dfrac{1}{x}\right)^6$ find:

  • (a) the coefficient of $x^2$
  • (b) the constant term

Solution:

First, establish the general term and simplify the powers of $x$:

$$ T_{r+1} = \binom{6}{r} (2x)^{6-r} (x^{-1})^r = \binom{6}{r} 2^{6-r} \cdot x^{6-r} \cdot x^{-r} = \binom{6}{r} 2^{6-r} \cdot x^{6-2r} $$

  • (a) For the coefficient of $x^2$:
    Set the power of $x$ to $2$:

    $$6 - 2r = 2 \implies 2r = 4 \implies r = 2$$

    Substitute $r=2$ into the simplified coefficient part:

    $$\text{Coefficient} = \binom{6}{2} 2^{6-2} = 15 \cdot 16 = 240$$

  • (b) For the constant term:
    The constant term has no $x$, which means the power of $x$ is $0$:

    $$6 - 2r = 0 \implies 2r = 6 \implies r = 3$$

    Substitute $r=3$ into the coefficient part:

    $$\text{Constant term} = \binom{6}{3} 2^{6-3} = 20 \cdot 8 = 160$$

Example 7

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

Solution:

Establish the general term and group the variables:

$$\begin{aligned} T_{r+1} &= \binom{12}{r} (2x^2)^{12-r} (-3x^{-1})^r \\ &= \binom{12}{r} 2^{12-r} \cdot x^{24-2r} \cdot (-3)^r \cdot x^{-r} \\ &= \binom{12}{r} 2^{12-r} (-3)^r \cdot x^{24-3r} \end{aligned}$$

For the constant term, set the exponent to $0$:

$$24 - 3r = 0 \implies 3r = 24 \implies r = 8$$

Substitute $r=8$:

$$ \text{Term} = \binom{12}{8} 2^{12-8} (-3)^8 = 495 \cdot (16) \cdot (6561) = 51963120 $$

Example 8 (Composite Expressions)

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

Solution:

The full expansion involves distributing the $(2x+3)$ across the expansion of $(4x+1)^7$. To create an $x^5$ term, we must multiply the $2x$ by the $x^4$ term, and the $3$ by the $x^5$ term.

  • 1. Find the $x^4$ term in $(4x+1)^7$: For $x^4$, we need $(4x)^4(1)^3$, so $r=3$.

    $$T_4 = \binom{7}{3}(4x)^4(1)^3 = 35 \cdot 256x^4 = 8960x^4$$

  • 2. Find the $x^5$ term in $(4x+1)^7$: For $x^5$, we need $(4x)^5(1)^2$, so $r=2$.

    $$T_3 = \binom{7}{2}(4x)^5(1)^2 = 21 \cdot 1024x^5 = 21504x^5$$

Now multiply by the respective parts of the first bracket:

$$\begin{aligned} \text{Total } x^5 \text{ term} &= 2x(8960x^4) + 3(21504x^5) \\ &= 17920x^5 + 64512x^5 \\ &= 82432x^5 \end{aligned}$$

Example 9HL Only

  • (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) Expand the right-hand side to verify:

    $$(2x+1)(x-2) = 2x^2 - 4x + x - 2 = 2x^2 - 3x - 2$$

  • (b) Using the factorization, we are looking for the coefficient of $x^2$ in:

    $$(2x+1)^5(x-2)^5$$

    Expand each factor ascendingly up to the $x^2$ term (higher powers will not contribute to the final $x^2$ term):

    First bracket:

    $$\begin{aligned} (1+2x)^5 &= 1^5 + \binom{5}{1}(1)^4(2x) + \binom{5}{2}(1)^3(2x)^2 + \dots \\ &= 1 + 5(2x) + 10(4x^2) + \dots \\ &= 1 + 10x + 40x^2 + \dots \end{aligned}$$

    Second bracket:

    $$\begin{aligned} (-2+x)^5 &= (-2)^5 + \binom{5}{1}(-2)^4(x) + \binom{5}{2}(-2)^3(x)^2 + \dots \\ &= -32 + 5(16)(x) + 10(-8)(x^2) + \dots \\ &= -32 + 80x - 80x^2 + \dots \end{aligned}$$

    Now multiply the expansions together, keeping only the cross-multiplications that result in $x^2$:

    $$(1 + 10x + 40x^2)(-32 + 80x - 80x^2)$$

    $$\begin{aligned} x^2 \text{ terms} &= 1(-80x^2) + (10x)(80x) + (40x^2)(-32) \\ &= -80x^2 + 800x^2 - 1280x^2 \\ &= -560x^2 \end{aligned}$$

    The coefficient is -560.