# Binomial theorem *Pascal's triangle* $$(a + b)^0 = 1$$ $$(a + b)^1 = a + b$$ $$(a + b)^2 = a^2 + 2ab + b^2$$ $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$

*Arranging the coefficients* The arrangement given is called the Pascal's triangle ## Examples 1. Give an expanded form of $(a + b)^4$ Taking the first three terms of the expansion find the value of the $(1.025)^4$ correct to 3 decimal places **Solution** From Pascal's triangle the coefficients are 1, 4, 6, 4, 1 $$(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$ For $(1.025)^4$, let $1.025 = 1 + \frac{1}{4}x$, so $x = 0.1$ $$(1 + 0.025)^4 \approx 1 + 4(0.025) + 6(0.025)^2 = 1 + 0.1 + 0.00375 = 1.10375 \approx 1.104$$ 2. Expand $(2 - x)^6$ in ascending powers of $x$. Taking $x = 0.002$ and using the first three terms of the expansion find the value of $(1.9998)^6$ as accurately as you can. Examine the fourth term of the expansion to find to how many places of decimals your answer is correctly **Solution** Coefficients = 1, 6, 15, 20, 15, 6, 1 $$ \begin{aligned} (2 - x)^6 &= 2^6 + 6(2)^5(-x) + 15(2)^4(-x)^2 + 20(2)^3(-x)^3 + 15(2)^2(-x)^4 + 6(2)(-x)^5 + (x)^6 \\ &= 64 - 192x + 240x^2 - 160x^3 + \dots \end{aligned} $$ putting $x = 0.002$ in the first three terms $$ \Rightarrow (2 - x)^6 \approx 64 - 0.384 + 0.000096 = 63.616096 $$ 3. Expand $(1 - 2x)^5$ in ascending powers of $x$ hence find $(0.98)^4$ to four decimal places. ## The Binomial Theorem If $n$ is a positive integer $$(a + b)^n = {}^nC_0 a^n + {}^nC_1 a^{n-1}b + {}^nC_2 a^{n-2}b^2 + \cdots + {}^nC_r a^{n-r}b^r + \cdots + {}^nC_n b^n$$ Where ${}^nC_r = \frac{n!}{r!(n-r)!}$ $n$ things taken $r$ at a time $$ \begin{aligned} 5! &= 5 \times 4 \times 3 \times 2 \times 1 = 120 \\ {}^nC_1 &= \frac{n!}{(n-1)!} = \frac{n(n-1)(n-2)...(1)}{(n-1)(n-2)...(1) \times 1} = n \\ {}^nC_2 &= \frac{n!}{(n-2)!2!} = \frac{n(n-1)(n-2)...(1)}{(n-2)(n-3)(n-4)...(1) \times 2!} = \frac{n(n-1)}{2!} \\ {}^nC_3 &= \frac{n!}{(n-3)!3!} = \frac{n(n-1)(n-2)(n-3)...(1)}{(n-3)(n-4)(n-5)...(1) \times 3!} = \frac{n(n-1)(n-2)}{3!} \\ \text{Hence} \\ \Rightarrow (a + b)^n &= a^n + na^{n-1}b + \frac{n(n-1)a^{n-2}b^2}{2!} + \frac{n(n-1)(n-2)a^{n-3}b^3}{3!} + \cdots + b^n \end{aligned} $$ ## Examples 1. Write down the term in $x^7$ in the expansion of $(2x - \frac{1}{2})^{12}$ **Solution** Coefficient $^{12}C_5 = \frac{12!}{5!7!} = 792$ The term is $$ \begin{aligned} &792(2x)^{12-5}\left(-\frac{1}{2}\right)^5 \\ &-792 \times 4x^7 \\ &= -3168x^7 \end{aligned} $$ 2. Write down the first 4 terms of the expansion of $\left(1-\frac{1}{2}x\right)^9$ in ascending powers of $x$ **Solution** $$ \left(1-\frac{1}{2}x\right)^9 = 1^9 + {}^9C_1\left(\frac{-1}{2}x\right) + {}^9C_2\left(\frac{-1}{2}x\right)^2 + {}^9C_3\left(\frac{-1}{2}x\right)^3 + {}^9C_4\left(\frac{-1}{2}x\right)^4 + \ldots $$ $$ = 1 - \frac{9x}{2} + \frac{9 \times 8}{2!} \left(\frac{1}{2}x\right)^2 - \frac{9 \times 8 \times 7}{3!} \left(\frac{1}{2}x\right)^3 + \ldots $$ $$ = 1 - \frac{9x}{2} + 9x^2 - \frac{21}{2}x^3 + \ldots $$ $$ \therefore \left(1-\frac{1}{2}x\right)^9 = 1 - \frac{9x}{2} + 9x^2 - \frac{21}{2}x^3 + \cdots (\text{to four decimal places}) $$ 3. Give the constant term in the expansion of $\left(2x^2 + \frac{1}{16x^4}\right)^6$ **Solution** The required term is $^{6}C_4 (2x^2)^2$ $$ \begin{aligned} &\rightarrow {}^6C_4 (2x^2)^2 \left(\frac{1}{16x^4}\right)^2 \\ &= \frac{6!}{2!4!} \times 4x^4 \times \frac{1}{256x^8} \\ &= 15 \times 4 \times \frac{1}{256} \\ &= \frac{60}{256} \\ &= \frac{15}{64} \end{aligned} $$ 4. Find the ratio of the term in $x^5$ to the term in $x^6$ in the expansion of $(1 - 2x)^{10}$ ## Binomial Series If $\alpha$ is any rational number, then $$(1 + x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \frac{\alpha(\alpha - 1)(\alpha - 2)}{3!}x^3 + \cdots$$ **Note:** 1) If $\alpha = n$ where $n \in \mathbb{Z}^+$, then the series terminates at $x^n$ 2) If $\alpha$ is not a positive integer then the series is infinite and converges only when $|x| < 1$ ($\alpha$ is a rational number) ## Examples Expand $\sqrt{1+x}$ $$\sqrt{1+x} = (1+x)^{\frac{1}{2}} = 1 + \frac{1}{2}x + \frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}x^2 + \frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}x^3 = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \ldots$$ The expansion is valid for $|x| < 1$ or $-1 < x < 1$ Note: the expansion is valid for $|x| < 1$ To expand $(a + x)^\alpha$ will be $a^\alpha\left(1 + \frac{x}{a}\right)^\alpha$ **Expand** $\left(1+\frac{1}{2}x\right)^{-3}$ **Solution** $$ \left(1+\frac{1}{2}x\right)^{-3} = 1 + (-3)\left(\frac{x}{2}\right) + \frac{(-3)(-4)}{2!}\left(\frac{x}{2}\right)^2 + \frac{(-3)(-4)(-5)}{3!}\left(\frac{x}{2}\right)^3 + \cdots $$ $$ = 1 - \frac{3x}{2} + \frac{12x^2}{4} - \frac{60x^3}{8} = 1 + \frac{3}{2}x + 3x^2 + \frac{5}{4}x^3 $$ The expression is valid when $|x| < 2$ **Expand** $\frac{1}{(1+x)(1-2x)}$ **up to and including the term in** $x^3$ **Solution** $$ \frac{1}{(1+x)(1-2x)} = (1+x)^{-1}(1-2x)^{-1} $$ $$ = \left(1 - x + x^2 - x^3 + \cdots\right)\left(1 + 2x + 4x^2 + 8x^3 + \cdots\right) $$ $$ = 1 + x + x^2 + 5x^3 + \cdots $$ The expansion is valid when $|x| < 1$ or $-1 < x < 1$ and $|2x| < 1$ or $-\frac{1}{2} < x < \frac{1}{2}$ $$ \text{i.e. } -\frac{1}{2} < x < \frac{1}{2} $$ **Alternative method using partial fractions:** $$ \begin{aligned} \frac{1}{(1+x)(1-2x)} &= \frac{A}{1+x} + \frac{B}{1-2x} \\ \text{when } x = \frac{1}{2} &\rightarrow 1 = \frac{3B}{2}, B = \frac{2}{3} \\ \text{when } x = -1 &\rightarrow 1 = 3A \rightarrow A = \frac{1}{3} \end{aligned} $$ $$ \begin{aligned} \frac{1}{(1+x)(1-2x)} &= \frac{1}{3(1+x)} + \frac{2}{3(1-2x)} \\ &= \frac{1}{3}(1+x)^{-1} + \frac{2}{3}(1-2x)^{-1} \\ &= \frac{1}{3}\left(1 - x + x^2 - x^3 + \cdots + 2\left(1 + 2x + 4x^2 + 8x^3 + \cdots\right)\right) \\ &= \frac{1}{3}(3 + 3x + 9x^2 + 15x^3 + \cdots) \\ &= 1 + x + 3x^2 + 5x^3 + \cdots \end{aligned} $$