Binomal Expansion

Expanding Binomials

Example 1

$$\begin{align} (x+2y)^3 &= 1(x)^3(2y)^0 + 3(x)^2(2y)^1 + 3(x)^1(2y)^2 + 1(x)^0(2y)^3 \\ &= x^3 + 6x^2y + 12xy^2 + 8y^3 \end{align}$$

Example 2

The coefficient of $x^2$ in the expansion of $(2-cx)^3$ is $294$. Find the possible value(s) of the constant $c$

$$\begin{align} (2-cx)^3 &\rightarrow 294x^2 \\ 3(2)(-cx)^2 &= 294x^2 \\ 6c^2x^2 &= 294x^2 \\ 6c^2 &= 294 \\ c^2 &= 49 \\ c &= \pm 7 \end{align}$$

Example 3

Expand $(4+k)^3$. Hence or otherwise, write down the expansion of $(4+x^2-x)^3$ in ascending powers of x.

Step 1: Expand

$$\begin{align} (4+k)^{3} &= 1(4)^{5}(k)^{0} + 3(4)^{1}(k)^{2} + 1(4)^{0}(k)^{3} \\ &= 64 + 48k + 12k^{2} + k^{3} \end{align}$$

Step 2: Substitute

$$\begin{align} &\text{Let }k = x^{2} - x\\ \\ &\implies 64 + 48(x^{2} - x) + 12(x^{2} - x)^{2} + (x^{2} - x)^{3} \\ &= 64 + 48(x^2 - x) + 12(x^4 - 2x^3 + x^2) + (x^6 - 3x^5 + 3x^4 - x^3) \\ &= 64 + 48x^{2} - 48x + 12x^{4} - 24x^{3} + 12x^{2} + x^{6} - 3x^{5} + 3x^{4} - x^{3} \\ &= x^{6} - 3x^{5} + 15x^{4} - 25x^{3} + 60x^{2} - 48x + 64 \end{align}$$

Choose

{}^nC_{r} = {n \choose r} = \frac{n!}{r! (n-r)!}

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Basic examples

$$\begin{align} \binom{15}{6} &= 5005 \\ \binom{10}{7} &= 600600 \end{align}$$ $$\begin{align} (3-2x)^{5} &= {10 \choose 0}(1)^{10}(2x)^{0} + {10 \choose 1}(1)^{9}(2x)^{1} + {10 \choose 2}(1)^{8}(2x)^{2} + \dots \\ &= 1(1)^{10}(2x)^{0} + 10(1)^{9}(2x)^{1} + 45(1)^{8}(2x)^{2} \\ &= 1 + 20x + 180x^{2} + 960x^{3} + \dots \end{align}$$

Given that ${50 \choose 13} = \frac{50!}{13!a!}$, write down the value of $a$

$$\begin{align} {n \choose r} &= \frac{n!}{r!(n-r)!} \\ {50 \choose 13} &= \frac{50!}{13!a!} \\[6pt] \implies a &= (n-r) \\ a &= (50 - 13) \\ a &= \boxed{ 37 } \end{align}$$

Given that ${35 \choose p} = \frac{35!}{p!18!}$, write down the value of $p$

$$\begin{align} {n \choose r} &= \frac{n!}{r!(n-r)!} \\ {35 \choose p} &= \frac{35!}{p!18!} \\[6pt] n &= 35 \\ r &= p \\ (n-r) &= 18 \\ (35 - p) &= 18 \\[6pt] -p &= -17 \\ p &= \boxed{ 17 } \end{align}$$

Prove $\binom{n}{1} = n$

$$\begin{align} \binom{n}{1} &= \frac{n!}{r!(n-r)!} \\ &= \frac{n!}{1 (n-1)!} \\ &= \frac{n!}{(n-1)!} \\ &= \frac{n \cancel{ (n-1) } \cancel{ (n-2) } \dots}{\cancel{ (n-1) } \cancel{ (n-2) } \dots} \\ &= n \end{align}$$

Prove ${n \choose 2} = \frac{n(n-1)}{2}$

$$\begin{align} {n \choose 2} &= \frac{n!}{r!(n-r)!} \\ &= \frac{n!}{2!(n-2)!} \\ &= \frac{n (n-1) \cancel{ (n-2) } \cancel{ (n-3) } \dots}{2 \cancel{ (n-2) } \cancel{ (n-3) }\dots} \\ &= \frac{n(n-1)}{2} \end{align}$$

Binomial Estimation

Find the first three terms in the expansion of $(1+2x)^{10}$

$$\begin{align} (1+2x)^{10} &= 1 + 10(2x) + \frac{10 \cdot 9}{2!}(2x)^2 + \dots \\ &= 1 + 20x + 45 \cdot 4x^2 + \dots \\ &= 1 + 20x + 180x^2 + \dots \end{align}$$

Use your expansion to approximate $1.20^{10}$. Give your answer to an appropriate number of decimal places.

$$\begin{align} 1+2x &= 1.02 \\ 2x &= 0.02 \\ x &= 0.01 \\ \\ (1.02)^{10} &= 1 + 20 \cdot 0.01 + 180 \cdot 0.0001 + \dots \\ &= 1 + 0.3 + 0.018 + \dots \\ &= 1.22\text{ (2 d.p.)} \end{align}$$