Look at the first few powers:
| $n$ | $(a+b)^n$ | Coefficients |
|---|---|---|
| $0$ | $1$ | $1$ |
| $1$ | $a + b$ | $1 \quad 1$ |
| $2$ | $a^2 + 2ab + b^2$ | $1 \quad 2 \quad 1$ |
| $3$ | $a^3 + 3a^2b + 3ab^2 + b^3$ | $1 \quad 3 \quad 3 \quad 1$ |
| $4$ | $a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$ | $1 \quad 4 \quad 6 \quad 4 \quad 1$ |
| $5$ | $\ldots$ | $1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1$ |
| $6$ | $\ldots$ | $1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1$ |
Expand $(1 + x)^4$.
For larger $n$, writing out Pascal's triangle is impractical. Instead each coefficient can be calculated directly.
Evaluate $5!$ and $\dbinom{7}{3}$.
| Useful fact | Example |
|---|---|
| $\dbinom{n}{0} = \dbinom{n}{n} = 1$ | Every row starts and ends with $1$ |
| $\dbinom{n}{1} = n$ | $\dbinom{8}{1} = 8$ |
| $\dbinom{n}{2} = \dfrac{n(n-1)}{2}$ | $\dbinom{6}{2} = 15$ |
| $\dbinom{n}{r} = \dbinom{n}{n-r}$ | $\dbinom{9}{2} = \dbinom{9}{7} = 36$ — the rows are symmetrical |
Expand $(2 - 3x)^5$ fully.
Find the coefficient of $x^3$ in the expansion of $(3 + 2x)^7$.
The coefficient of $x^2$ in the expansion of $(1 + kx)^8$ is $112$. Find the possible values of $k$.
Find the term independent of $x$ in the expansion of $\left(x + \dfrac{2}{x}\right)^6$.
Find the coefficient of $x^2$ in the expansion of $(1 + 2x)(1 - x)^6$.
(a) Find the first four terms, in ascending powers of $x$, of $\left(1 - \dfrac{x}{2}\right)^{10}$. (b) Use a suitable value of $x$ to estimate $0.99^{10}$.
Pascal's triangle
Each entry is the sum of the two above it.
Factorial
$n! = n(n-1)\cdots 1$; $0! = 1$.
$^n\mathrm{C}_r$
$\dfrac{n!}{r!(n-r)!}$; row $n$, position $r$ (from $0$).
Number of terms
$(a+b)^n$ has $n+1$ terms.
Powers
In every term the powers of $a$ and $b$ add to $n$.
General term
$\dbinom{n}{r}a^{n-r}b^r$
$(1+x)^n$
$1 + nx + \dfrac{n(n-1)}{2}x^2 + \cdots$
Brackets
Raise the whole of $(-3x)$ to the power, sign included.
Approximating
Choose $x$ small; a few terms suffice.
Check
Substitute $x = 1$ into both sides.
Use Pascal's triangle to expand $(1 + x)^5$.
▶ Show solution
Row 5: $1, 5, 10, 10, 5, 1$.
$(1 + x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$
Without a calculator, evaluate (a) $\dbinom{8}{3}$ (b) $\dbinom{10}{8}$.
▶ Show solution
(a) $\dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = \dfrac{336}{6} = 56$
(b) By symmetry $\dbinom{10}{8} = \dbinom{10}{2} = \dfrac{10 \times 9}{2} = 45$.
Expand $(2 + x)^4$ fully.
▶ Show solution
Row 4: $1, 4, 6, 4, 1$.
$2^4 + 4(2^3)x + 6(2^2)x^2 + 4(2)x^3 + x^4$
$= 16 + 32x + 24x^2 + 8x^3 + x^4$
Find the first three terms, in ascending powers of $x$, of $(1 - 3x)^9$.
▶ Show solution
$1 + 9(-3x) + \dbinom{9}{2}(-3x)^2$
$= 1 - 27x + 36(9x^2)$
$= 1 - 27x + 324x^2$
Find the coefficient of $x^4$ in the expansion of $(2x - 1)^7$.
▶ Show solution
Take $a = 2x$, $b = -1$. The $x^4$ term needs $a^4$, so the power of $b$ is $3$.
$\dbinom{7}{3}(2x)^4(-1)^3 = 35 \times 16x^4 \times (-1)$
Coefficient $= -560$.
The first three terms of $(1 + ax)^n$, in ascending powers of $x$, are $1 + 20x + 180x^2$. Find $a$ and $n$.
▶ Show solution
Compare with $1 + n(ax) + \dfrac{n(n-1)}{2}(ax)^2$:
$na = 20$ and $\dfrac{n(n-1)a^2}{2} = 180$.
Write the second as $\dfrac{na \times (n-1)a}{2} = 180$, so $\dfrac{20(n-1)a}{2} = 180$ and $(n-1)a = 18$.
That is $na - a = 18$, so $20 - a = 18$ and $a = 2$. Then $n = 10$.
Check: $\dbinom{10}{2}(2)^2 = 45 \times 4 = 180$ ✓
(a) Find the first three terms, in ascending powers of $x$, of $\left(1 + \dfrac{x}{4}\right)^8$. (b) Hence estimate $1.0025^8$, stating the value of $x$ you use.
▶ Show solution
(a) $1 + 8\left(\tfrac{x}{4}\right) + 28\left(\tfrac{x}{4}\right)^2 = 1 + 2x + \tfrac{7}{4}x^2$
(b) $1 + \tfrac{x}{4} = 1.0025$ gives $x = 0.01$.
$1 + 0.02 + \tfrac{7}{4}(0.0001) = 1.020175$
(The true value is $1.0201759\ldots$, so the estimate is accurate to six decimal places.)
Find the term independent of $x$ in the expansion of $\left(x^2 + \dfrac{2}{x}\right)^6$.
▶ Show solution
General term: $\dbinom{6}{r}\left(x^2\right)^{6-r}\left(\dfrac{2}{x}\right)^r = \dbinom{6}{r}2^r x^{12 - 2r - r} = \dbinom{6}{r}2^r x^{12 - 3r}$.
Independent of $x$: $12 - 3r = 0$, so $r = 4$.
$\dbinom{6}{4}2^4 = 15 \times 16 = 240$
Find the coefficient of $x^2$ in the expansion of $(2 - x)(1 + 3x)^5$.
▶ Show solution
$(1 + 3x)^5 = 1 + 5(3x) + 10(3x)^2 + \cdots = 1 + 15x + 90x^2 + \cdots$
$x^2$ terms: $2 \times 90x^2$ and $(-x) \times 15x$.
Coefficient $= 180 - 15 = 165$.
(a) Expand $(2 + x)^5$ fully, in ascending powers of $x$.
(b) Hence show that $(2 + x)^5 - (2 - x)^5 = 160x + 80x^3 + 2x^5$.
(c) Hence find the exact value of $\left(2 + \sqrt{3}\right)^5 - \left(2 - \sqrt{3}\right)^5$.
(d) Use the first three terms of your expansion in (a) to estimate $2.01^5$.
▶ Show solution
(a) Row 5: $1, 5, 10, 10, 5, 1$.
$32 + 5(16)x + 10(8)x^2 + 10(4)x^3 + 5(2)x^4 + x^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5$
(b) Replacing $x$ by $-x$ changes the sign of the odd powers only:
$(2 - x)^5 = 32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5$
Subtracting, the even powers cancel and the odd powers double: $160x + 80x^3 + 2x^5$ ✓
(c) Put $x = \sqrt{3}$: $\left(\sqrt{3}\right)^3 = 3\sqrt{3}$ and $\left(\sqrt{3}\right)^5 = 9\sqrt{3}$.
$160\sqrt{3} + 80(3\sqrt{3}) + 2(9\sqrt{3}) = 160\sqrt{3} + 240\sqrt{3} + 18\sqrt{3} = 418\sqrt{3}$
(d) $x = 0.01$: $32 + 80(0.01) + 80(0.0001) = 32 + 0.8 + 0.008 = 32.808$
(The true value is $32.808040\ldots$)