Series

1.6 Series

The binomial expansion

For a positive integer $n$:

$(a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + b^n$

where $\dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$. The term containing $b^r$ is

$\binom{5}{2} \times 2^3 \times x^2 = 10 \times 8 \times x^2 = 80x^2$

The coefficient is 80. Put any negative sign or coefficient of $x$ inside the bracket: in $(2 - 3x)^5$, the term has $(-3x)^2 = 9x^2$.

Finding a particular term

When the terms include powers of $x$ in both parts, write the general term and collect the powers.

The general term is $\dbinom{6}{r}x^{6-r}\left(\dfrac{2}{x}\right)^r = \dbinom{6}{r}2^r x^{6-2r}$.

"Independent of $x$" means the power is zero, so $r = 3$, and the term is $20 \times 8 = 160$.

Multiplying by another bracket

$(2 + x)^5 = 32 + 80x + 80x^2 + \dots$

The $x^2$ terms come from $1 \times 80x^2$ and $-2x \times 80x$. So the coefficient is $80 - 160 = -80$.

Arithmetic progressions

An arithmetic progression (AP) adds a common difference $d$ each time. With first term $a$:

$a + 4d = 17$ and $a + 11d = 45$. Subtracting gives $7d = 28$, so $d = 4$ and $a = 1$.

$S_{20} = \frac{20}{2}\left[2 + 19 \times 4\right] = 10 \times 78 = 780$

Geometric progressions

A geometric progression (GP) multiplies by a common ratio $r$ each time:

$ar^2 = 12$ and $ar^5 = 96$. Dividing gives $r^3 = 8$, so $r = 2$ and $a = 3$.

$S_{10} = \frac{3(1 - 2^{10})}{1 - 2} = 3 \times 1023 = 3069$

Telling them apart

Three consecutive terms $x$, $y$, $z$ are:

• in an AP if $y - x = z - y$
• in a GP if $\dfrac{y}{x} = \dfrac{z}{y}$

$\frac{k + 6}{k} = \frac{3k}{k + 6} \quad\Rightarrow\quad (k + 6)^2 = 3k^2 \quad\Rightarrow\quad k^2 - 6k - 18 = 0$

$k = \dfrac{6 + \sqrt{108}}{2} = 3 + 3\sqrt{3}$, taking the positive root.

Sum to infinity

When $|r| < 1$, the powers of $r$ shrink towards zero, so the sum settles at a limit. The GP is convergent.

$\dfrac{5}{1 - r} = 20$, so $1 - r = \dfrac{1}{4}$ and $r = \dfrac{3}{4}$.

Common forms of question

• Mixed AP and GP. Set up one equation from each, then solve them together. Reject a value of $r$ with $|r| \geq 1$ when a sum to infinity exists.
• Least value of $n$. Form an inequality from the sum, then try values of $n$ or use logarithms. Dividing by $\ln r$ when $r < 1$ reverses the inequality, because that logarithm is negative.
• Real contexts. Savings or lengths growing by a fixed amount are an AP. Growth by a fixed percentage is a GP with $r = 1 + \dfrac{p}{100}$.