Expand a few powers of $(a+b)$ by hand and the coefficients form a triangle, each number the sum of the two above it.
nCr key rather than writing out fifteen rows.
Row $n$ of Pascal's triangle is exactly $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$.
Evaluate $\binom{7}{3}$.
Evaluate $\binom{20}{18}$.
Expand $(1 + x)^5$.
Expand $(3 + 2x)^5$ in ascending powers of $x$.
Expand $(x - 2)^4$.
Questions often ask only for the coefficient of a particular power. There is no need to write the whole expansion.
- Decide which value of $r$ gives the power you want.
- Write down $\binom{n}{r}$.
- Raise each part of $a$ and $b$ to its power — including the numbers.
- Multiply everything together.
Find the coefficient of $x^4$ in the expansion of $(2 - 3x)^7$.
Use the expansion of $(1+x)^6$ to estimate $1.02^6$, and compare with the exact value.
The theorem
$(a+b)^n = \sum \binom{n}{r}a^{n-r}b^r$.
Number of terms
$n+1$.
Powers
Always add to $n$.
Coefficients
Row $n$ of Pascal's triangle.
Quick $\binom{n}{r}$
$r$ descending factors over $r!$.
Symmetry
$\binom{n}{r} = \binom{n}{n-r}$.
Brackets
$(2x)^3 = 8x^3$ — the number gets cubed too.
Negative $b$
Signs alternate.
One term
$\binom{n}{r}a^{n-r}b^r$; pick $r$ from the power wanted.
Check
Set $x=1$: coefficients of $(1+x)^n$ sum to $2^n$.
Write down row $6$ of Pascal's triangle.
▶ Show solution
Row $5$ is $1, 5, 10, 10, 5, 1$. Adding adjacent pairs:
$1,\; 6,\; 15,\; 20,\; 15,\; 6,\; 1$
Seven entries, as expected for $n = 6$.
Evaluate $\binom{8}{3}$ and $\binom{8}{5}$. Comment on your answers.
▶ Show solution
$\binom{8}{3} = \dfrac{8 \times 7 \times 6}{6} = 56$
$\binom{8}{5} = \binom{8}{3} = 56$
They are equal, illustrating the symmetry $\binom{n}{r} = \binom{n}{n-r}$.
Expand $(1 + x)^6$.
▶ Show solution
$1 + 6x + 15x^2 + 20x^3 + 15x^4 + 6x^5 + x^6$
Expand $(2 + x)^4$.
▶ Show solution
Coefficients $1, 4, 6, 4, 1$ with $a = 2$:
$2^4 + 4(2^3)x + 6(2^2)x^2 + 4(2)x^3 + x^4$
$= 16 + 32x + 24x^2 + 8x^3 + x^4$
Expand $(1 - 2x)^4$.
▶ Show solution
$b = -2x$, coefficients $1, 4, 6, 4, 1$:
$1 + 4(-2x) + 6(-2x)^2 + 4(-2x)^3 + (-2x)^4$
$= 1 - 8x + 24x^2 - 32x^3 + 16x^4$
Find the coefficient of $x^3$ in the expansion of $(1 + x)^{10}$.
▶ Show solution
$\binom{10}{3} = \dfrac{10 \times 9 \times 8}{6} = 120$
Find the term in $x^2$ in the expansion of $(3 + x)^6$.
▶ Show solution
$r = 2$, so the term is $\binom{6}{2}(3)^4 x^2$.
$= 15 \times 81 \times x^2 = 1215x^2$
Find the coefficient of $x^3$ in the expansion of $(2x - 1)^5$.
▶ Show solution
Take $a = 2x$ and $b = -1$. The power of $x$ comes from $a$, so we need $a^3$, i.e. $n - r = 3$ and $r = 2$.
Term $= \binom{5}{2}(2x)^3(-1)^2 = 10 \times 8x^3 \times 1 = 80x^3$
The coefficient is $80$.
In the expansion of $(1 + kx)^5$ the coefficient of $x^2$ is $90$. Find the possible values of $k$.
▶ Show solution
The $x^2$ term is $\binom{5}{2}(kx)^2 = 10k^2x^2$.
$10k^2 = 90 \;\Rightarrow\; k^2 = 9$
$k = 3$ or $k = -3$.
Both work because the coefficient depends on $k^2$, which loses the sign. The $x$ coefficients would differ ($15$ and $-15$), but the question only fixes the $x^2$ term.
(a) Expand $(1 + x)^8$ as far as the term in $x^3$. (b) Use this to estimate $1.01^8$ to $6$ decimal places. (c) Find the coefficient of $x^5$ in $\left(x^2 + \tfrac{2}{x}\right)^4$. (d) Explain why the expansion in (a) would be a poor way to estimate $1.5^8$.
▶ Show solution
(a) Coefficients $\binom{8}{0} = 1$, $\binom{8}{1} = 8$, $\binom{8}{2} = 28$, $\binom{8}{3} = 56$:
$(1+x)^8 \approx 1 + 8x + 28x^2 + 56x^3$
(b) Put $x = 0.01$:
$1 + 0.08 + 28(0.0001) + 56(0.000001)$
$= 1 + 0.08 + 0.0028 + 0.000056 = 1.082856$
(The exact value is $1.0828567\ldots$, so this is correct to $6$ d.p.)
(c) Here $a = x^2$ and $b = \tfrac2x = 2x^{-1}$. The general term is
$\binom{4}{r}\left(x^2\right)^{4-r}\left(2x^{-1}\right)^r = \binom{4}{r}2^r x^{8-2r-r} = \binom{4}{r}2^r x^{8-3r}$
We need $8 - 3r = 5$, so $r = 1$.
Coefficient $= \binom{4}{1} \times 2^1 = 4 \times 2 = 8$
(d) Estimating $1.5^8$ needs $x = 0.5$, which is not small. The dropped terms are then large — the next one, $\binom{8}{4}x^4 = 70(0.0625) = 4.375$, is bigger than the $x^3$ term we kept.
The four-term estimate gives $1 + 4 + 7 + 7 = 19$, whereas $1.5^8 = 25.6$ — an error of about $26\%$. The approximation only works when $x$ is small enough that later terms shrink quickly.