4.2 Binomial expansion for rational n and approximation — Practise with help

1. $(1 + x)^3 = 1 + 3x + 3x^2 + x^3$.
2. $9^{\frac{1}{2}} = 3$ and $27^{\frac{1}{3}} = 3$, because $3^2 = 9$ and $3^3 = 27$.
3. $\left(\dfrac{x}{2}\right)^3 = \dfrac{x^3}{8}$.

If question 2 was hard, revisit fractional indices in Section 2.1 before you continue. A fractional power is a root.

Use $n = -1$ in the general series. Each coefficient $\binom{-1}{r}$ is $(-1)^r$, so the coefficients follow a simple pattern.

Work out the first three coefficients: $\binom{-1}{1} = -1$, $\binom{-1}{2} = \dfrac{(-1)(-2)}{2} = 1$, and $\binom{-1}{3} = \dfrac{(-1)(-2)(-3)}{6} = -1$.

With $n = -1$,

The signs alternate. This is a geometric series with first term $1$ and ratio $-x$.

Start from the answer to question 1 and replace each $x$ with $3x$. The range changes too.

Work out the powers of $3x$: $(3x)^2 = 9x^2$ and $(3x)^3 = 27x^3$. Then apply the coefficients $1, -1, 1, -1$.

The expansion is valid when $|3x| < 1$, which is $|x| < \dfrac{1}{3}$. The coefficients grow because each term brings another factor of $3$.

Use $n = \frac{1}{2}$ with $-x$ in place of $x$. You did the coefficients for $n = \frac{1}{2}$ in the Learn page.

Take $\binom{\frac{1}{2}}{2} = -\dfrac{1}{8}$ and $\binom{\frac{1}{2}}{3} = \dfrac{1}{16}$, then attach $(-x)^2 = x^2$ and $(-x)^3 = -x^3$.

The odd powers are still negative, because the second factor is $-\frac{1}{2}$. The even power $(-x)^2 = x^2$ is positive but carries the coefficient $-\frac{1}{8}$.

Take out the factor $9^{\frac{1}{2}} = 3$ first, so the bracket is $\left(1 + \dfrac{x}{9}\right)^{\frac{1}{2}}$.

Use the coefficients $1, \frac{1}{2}, -\frac{1}{8}, \frac{1}{16}$ with $\dfrac{x}{9}$ in place of $x$, then multiply every term by $3$.

The expansion is valid when $\left|\dfrac{x}{9}\right| < 1$, which is $|x| < 9$.

Write $1.02 = 1 + 0.02$, so $x = 0.02$. Take the terms up to $x^2$.

The first three terms are $1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2$. Substitute $x = 0.02$, so the middle term is $0.01$ and the last term is $-\dfrac{1}{8}(0.0004)$.

A calculator gives $1.0099505$ to $7$ decimal places, so the three-term estimate is correct to $6$ decimal places. The term in $x^3$ is about $-0.0000005$, which is smaller than the rounding.

Write the expression as $(1 + 4x)^{-\frac{1}{2}}$, so $n = -\frac{1}{2}$ and the replacement for $x$ is $4x$.

Work out the coefficients $\binom{-\frac{1}{2}}{1} = -\dfrac{1}{2}$ and $\binom{-\frac{1}{2}}{2} = \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2} = \dfrac{3}{8}$. Then attach $4x$ and $(4x)^2 = 16x^2$.

The range of validity is $|x| < \dfrac{1}{4}$. The second term is $-\frac{1}{2} \times 4x = -2x$, and the third is $\frac{3}{8} \times 16x^2 = 6x^2$.