General Binomial Expansion (Edexcel A-Level Mathematics): Revision Notes

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

When expanding brackets of the form $(1 + x)^n$ where $n$ is not necessarily a positive integer, the standard formula can be used.

Example: Expand $(1 - x)^3$

Thus, the expansion is:

Example: Expand up to and including $x^3, (1 + 2x)^{-1}$

Example: Expand $\left(1 - \frac{1}{2}x\right)^{\frac{1}{2}}$ up to and including the $x^3$ term

The lottery jackpot is won by choosing 6 numbers correctly out of 49. What is the probability of winning the jackpot?

Calculation

• The number of ways to choose $6$ numbers out of $49$ is given by the binomial coefficient: $\dbinom{49}{6} = \dfrac{49!}{6!(49-6)!}$

• Calculating this, we get: $\dbinom{49}{6} = 13,983,816$

• Therefore, the probability of winning the jackpot is: $\dfrac{1}{13,983,816}$

Example: Expand $(2 + x)^6$ fully

1. General Term in Binomial Expansion: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

2. Apply to ($2 + x)^6$:

• Here $a = 2,\ b = x$, and $n = 6$.

3. Consider each term separately:

• Constant Term: $(2)^6 \times (x)^0 \times \binom{6}{0} = 64$

• $x^1$ Term: $(2)^5 \times (x)^1 \times \binom{6}{1} = 192x$

• $x^2$ Term: $(2)^4 \times (x)^2 \times \binom{6}{2} = 240x^2$

• $x^3$ Term: $(2)^3 \times (x)^3 \times \binom{6}{3} = 160x^3$

• $x^4$ Term: $(2)^2 \times (x)^4 \times \binom{6}{4} = 60x^4$

• $x^5$ Term: $(2)^1 \times (x)^5 \times \binom{6}{5} = 12x^5$

• $x^6$ Term: $(2)^0 \times (x)^6 \times \binom{6}{6} = x^6$

4. Combine all terms: $(2 + x)^6 = 64 + 192x + 240x^2 + 160x^3 + 60x^4 + 12x^5 + x^6$

Example 1: Expand $(3 + 2x)^4$

5. General Term in Binomial Expansion: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

6. Apply to $(3 + 2x)^4$:

• Here $a = 3,\ b = 2x,\ and\ n = 4$.

7. Calculate each term: | ${}^4C_0$ | ${}^4C_1$ | ${}^4C_2$ | ${}^4C_3$ | ${}^4C_4$ | |---|---|---|---|---| | $(3)^4$ | $(3)^3$ | $(3)^2$ | $(3)^1$ | $(3)^0$ | | $(2x)^0$ | $(2x)^1$ | $(2x)^2$ | $(2x)^3$ | $(2x)^4$ |

• $\binom{4}{0}\cdot(3)^4\cdot(2x)^0 = 1 \cdot 81 \cdot 1 = 81$
• $\binom{4}{1}(3)^3(2x)^1 = 4 \cdot 27 \cdot 2x = 216x$
• $\binom{4}{2}(3)^2(2x)^2 = 6 \cdot 9 \cdot 4x^2 = 216x^2$
• $\binom{4}{3}(3)^1(2x)^3 = 4 \cdot 3 \cdot 8x^3 = 96x^3$
• $\binom{4}{4}(3)^0(2x)^4 = 1 \cdot 1 \cdot 16x^4 = 16x^4$

8. Combine all terms: $(3 + 2x)^4 = 81 + 216x + 216x^2 + 96x^3 + 16x^4$

Example 2: Expand $(6 - 2x)^5$

1. Apply to $(6 - 2x)^5$:

• Here $a = 6,\ b = -2x,\ and\ n = 5$.

2. Calculate each term: | ${}^5C_0$ | ${}^5C_1$ | ${}^5C_2$ | ${}^5C_3$ | ${}^5C_4$ | ${}^5C_5$ | |---|---|---|---|---|---| | $(6)^5$ | $(6)^4$ | $(6)^3$ | $(6)^2$ | $(6)^1$ | $(6)^0$ | | $(-2x)^0$ | $(-2x)^1$ | $(-2x)^2$ | $(-2x)^3$ | $(-2x)^4$ | $(-2x)^5$ |

• $\binom{5}{0}\cdot(6)^5\cdot(-2x)^0 = 1 \cdot 7776 \cdot 1 = 7776$
• $\binom{5}{1} \cdot (6)^4 \cdot (-2x)^1 = 5 \cdot 1296 \cdot -2x = -12960x$
• $\binom{5}{2}\cdot(6)^3\cdot(-2x)^2 = 10 \cdot 216 \cdot 4x^2 = 8640x^2$
• $\binom{5}{3}\cdot(6)^2\cdot(-2x)^3 = 10 \cdot 36 \cdot -8x^3 = -2880x^3$
• $\binom{5}{4}\cdot(6)^1\cdot(-2x)^4 = 5 \cdot 6 \cdot 16x^4 = 480x^4$
• $\binom{5}{5}\cdot(6)^0\cdot(-2x)^5 = 1 \cdot 1 \cdot -32x^5 = -32x^5$

3. Combine all terms: $(6 - 2x)^5 = 7776 - 12960x + 8640x^2 - 2880x^3 + 480x^4 - 32x^5$

Example 1: Expand $(1 - \frac{1}{3}x)^4$

1. Apply to $(1 - \frac{1}{3}x)^4$:

• Here $a = 1,\ b = -\frac{1}{3}x,\ and\ n = 4$.

2. Calculate each term: | ${}^4C_4$ | ${}^4C_3$ | ${}^4C_2$ | ${}^4C_1$ | ${}^4C_0$ | |---|---|---|---|---| | $1^4$ | $1^3$ | $1^2$ | $1^1$ | $1^0$ | | $(-\frac{1}{3}x)^0$ | $(-\frac{1}{3}x)^1$ | $(-\frac{1}{3}x)^2$ | $(-\frac{1}{3}x)^3$ | $(-\frac{1}{3}x)^4$ |

• $\binom{4}{0}\cdot(1)^4\cdot(-\frac{1}{3}x)^0 = 1 \cdot 1 \cdot 1 = 1$
• $\binom{4}{1}\cdot(1)^3\cdot(-\frac{1}{3}x)^1 = 4 \cdot 1 \cdot -\frac{1}{3}x = -\frac{4}{3}x$
• $\binom{4}{2}\cdot(1)^2\cdot(-\frac{1}{3}x)^2 = 6 \cdot 1 \cdot \frac{1}{9}x^2 = \frac{2}{3}x^2$
• $\binom{4}{3}\cdot(1)^1\cdot(-\frac{1}{3}x)^3 = 4 \cdot 1 \cdot -\frac{1}{27}x^3 = -\frac{4}{27}x^3$
• $\binom{4}{4}\cdot(1)^0\cdot(-\frac{1}{3}x)^4 = 1 \cdot 1 \cdot \frac{1}{81}x^4 = \frac{1}{81}x^4$

3. Combine all terms: $(1 - \frac{1}{3}x)^4 = 1 - \frac{4}{3}x + \frac{2}{3}x^2 - \frac{4}{27}x^3 + \frac{1}{81}x^4$

Example: Find the coefficient of the $x^6$ term in $(2 + x)^{12}$

1. General Term: $\text{Term} = \binom{12}{k} (2)^{12-k} (x)^k$

2. For $x^6$ term: $k = 6$

$\text{Term} = \binom{12}{6} (2)^{12-6} (x)^6 = \binom{12}{6} (2)^6 x^6$

3. Calculate the coefficient: $\binom{12}{6} = 924$

$(2)^6 = 64$

$\text{Coefficient} = 924 \times 64 = 59136$

So, the coefficient of the $x^6$ term is 59136.

Example: Expand $(3 + \frac{1}{2}x)^6$ up to and including the $x^2$ term, in ascending powers of x.

1. General Term: $\text{Term} = \binom{6}{k} (3)^{6-k} \left(\frac{1}{2}x\right)^k$

2. Calculate each term up to $x^2$:

• Constant term $(x^0$): $\binom{6}{0}\cdot (3)^6 \cdot\left(\frac{1}{2}x\right)^0 = 1 \cdot 729 \cdot 1 = 729$

• $x^1$ term: $\binom{6}{1}\cdot (3)^5 \cdot\left(\frac{1}{2}x\right)^1 = 6 \cdot 243 \cdot \frac{1}{2}x = 729x$

• $x^2$ term: $\binom{6}{2} \cdot(3)^4 \cdot\left(\frac{1}{2}x\right)^2 = 15 \cdot 81 \cdot \frac{1}{4}x^2 = 304.5x^2$

3. Combine these terms: $(3 + \frac{1}{2}x)^6 \approx 729 + 729x + \frac {1215}{4}x^2+...$

Example: Expand $(1 + 2x)^6$ in ascending powers of x up to the $x^2$ term. Use this expansion to approximate $1.04^6$.

Step 1: Expand $(1 + 2x)^6$

1. General Term in Binomial Expansion: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

2. Apply to $(1 + 2x)^6$:

• Here $a = 1,\ b = 2x,\ and\ n = 6.$

3. Calculate each term up to $x^2$:

• Constant term $(x^0)$: $\binom{6}{0}\cdot(1)^6\cdot(2x)^0 = 1$

• $x^1$ term: $\binom{6}{1}\cdot(1)^5\cdot(2x)^1 = 6 \cdot 1 \cdot 2x = 12x$

• $x^2$ term: $\binom{6}{2}\cdot(1)^4\cdot(2x)^2 = 15 \cdot 1 \cdot 4x^2 = 60x^2$

4. Combine these terms: $(1 + 2x)^6 \approx 1 + 12x + 60x^2 + \dots$

Step 2: Use this expansion to approximate $1.04^6$

1. Set up the approximation:

• Let $1 + 2x = 1.04.$

2. Solve for $x$:

• $1 + 2x = 1.04$
• $2x = 0.04$
• $x = 0.02$

3. Substitute $x = 0.02$ into the expansion: $1.04^6 \approx 1 + 12(0.02) + 60(0.02)^2$

4. Calculate each term:

• $12(0.02) = 0.24$
• $60(0.02)^2 = 60(0.0004) = 0.024$

5. Combine these values: $1.04^6 \approx 1 + 0.24 + 0.024 = 1.264$

Final Approximation

$1.04^6 \approx :highlight[1.264]$

Example: Expand $\left( x + \frac{1}{x} \right)^4$

1. General Term in Binomial Expansion: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

2. Apply to $\left( x + \frac{1}{x} \right)^4$:

• Here $a = x,\ b = \frac{1}{x},\ and \ n = 4$.

3. Calculate each term:

• $\binom{4}{0}\cdot(x)^4\cdot\left(\frac{1}{x}\right)^0 = x^4$
• $\binom{4}{1}\cdot(x)^3\cdot\left(\frac{1}{x}\right)^1 = 4x^2$
• $\binom{4}{2}\cdot(x)^2\cdot\left(\frac{1}{x}\right)^2 = 6$
• $\binom{4}{3}\cdot(x)^1\cdot\left(\frac{1}{x}\right)^3 = \frac{4}{x^2}$
• $\binom{4}{4}\cdot(x)^0\cdot\left(\frac{1}{x}\right)^4 = \frac{1}{x^4}$

4. Combine all terms: $\left( x + \frac{1}{x} \right)^4 = x^4 + 4x^2 + 6 + \frac{4}{x^2} + \frac{1}{x^4}$

Example: Find the coefficient of $x^3$ in the expansion of $(2x)(3 + 2x)^6$

1. Use the binomial expansion for $(3 + 2x)^6$: $(3 + 2x)^6 = \sum_{k=0}^{6} \binom{6}{k} 3^{6-k} (2x)^k$

2. Identify the term involving $x^3$:

• For$(3 + 2x)^6$, the $x^3$ term is given by: $\binom{6}{3} 3^{6-3} (2x)^3 = \binom{6}{3} 27 (8x^3) = 20,160x^3$

3. Multiply by the $2x$ term:

• The final term involving $x^3$ in $(2x)(3 + 2x)^6$: $2x \cdot 20,160x^3 = 40,320x^4$

Example: Find the coefficient of $x^8$ in $(1 + 2x^2)^6$

1. Use the binomial expansion for $(1 + 2x^2)^6$: $(1 + 2x^2)^6 = \sum_{k=0}^{6} \binom{6}{k} 1^{6-k} (2x^2)^k$

2. Identify the term involving $x^8$:

• For $(1 + 2x^2)^6$, the $x^8$ term is given by: $\binom{6}{4} 1^{6-4} (2x^2)^4 = \binom{6}{4} 1^2 (16x^8) = 240x^8$