Sequences, series and the binomial expansion: A-level Maths OCR revision notes

1.04

Sequences and series

Sequences and series: the language 1.04

Definitions

Sequence: an ordered list of terms $u_{1}, u_{2}, u_{3}, \dots$ , where $u_{n}$ denotes the $n$ th term.
Series: the sum of the terms of a sequence; $S_{n}$ denotes the sum of the first $n$ terms.
Position-to-term rule: a formula giving $u_{n}$ directly from $n$ , as opposed to a recurrence relation such as $u_{n + 1} = u_{n} + 4$ , which gives each term from the previous one.

Notes

A recurrence relation needs a stated first term to define the sequence; a position-to-term rule does not.
Throughout this topic $a$ denotes the first term, $d$ the common difference of an arithmetic sequence, $r$ the common ratio of a geometric one, and $l$ the last term of a finite series.
$u_{n}$ and $S_{n}$ are easily confused under time pressure — $u_{n}$ is a single term, $S_{n}$ is the running total of the first $n$ terms; check which one a question is actually asking for before substituting into a formula.

1.04

Arithmetic sequences and series

Arithmetic sequences 1.04

Definitions

Arithmetic sequence: one in which consecutive terms differ by a constant common difference $d$ , so $u_{n + 1} - u_{n} = d$ for every $n$ .

Key results

$n$ th term: $u_{n} = a + (n - 1) d$ .
The common difference is found from any two consecutive terms, or from two terms $p$ places apart as $d = \frac{u _{m + p} - u _{m}}{p}$ .

Notes

The coefficient of $n$ in the $n$ th term is always $d$ , so a sequence whose $n$ th term is linear in $n$ is arithmetic, and one whose $n$ th term is not linear is not.
A negative $d$ gives a decreasing sequence; the terms eventually become negative however large $a$ is.
' $n$ places apart' means the difference between the two subscripts, not the two term values — $u_{5}$ and $u_{9}$ are $4$ places apart, not $5$ .

Arithmetic series 1.04

Key results

Sum of the first $n$ terms: $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$ .
Equivalently, using the first term $a$ and the last term $l$ : $S_{n} = \frac{n}{2} (a + l)$ — the two forms are the same identity viewed two ways, since $l = a + (n - 1) d$ .
The sum of a block of terms from the $p$ th to the $q$ th inclusive is $S_{q} - S_{p - 1}$ , and that block contains $q - p + 1$ terms.

Method

Translate each piece of given information into an equation in $a$ and $d$ using $u_{n} = a + (n - 1) d$ or the sum formula.
Solve the resulting simultaneous equations for $a$ and $d$ .
Substitute back into whichever formula the question actually asks about, and sanity-check by listing the first few terms.

In practiceThe 5th term of an arithmetic series is 17 and the sum of the first 10 terms is 185. Find $a$ and $d$ .

$u_{5} = a + 4 d = 17$
$S_{10} = \frac{10}{2} (2 a + 9 d) = 185 ⟹ 2 a + 9 d = 37$
$(2 a + 9 d) - 2 (a + 4 d) = 37 - 34 ⟹ d = 3, a = 5$
$5, 8, 11, 14, 17, \dots$ check: the 5th term is 17

Notes

The sum formula comes from pairing the first term with the last, the second with the second-last and so on: each pair totals $a + l$ , and there are $\frac{n}{2}$ such pairs.
A common source of error is miscounting the number of terms in a sum, especially when a question specifies a range such as 'the 5th to the 20th term inclusive' — that range contains $20 - 5 + 1 = 16$ terms, not $15$ .
When $S_{n}$ is given and $n$ is unknown, the sum formula produces a quadratic in $n$ ; discard any negative or non-integer root, since $n$ counts terms.
Both forms of the sum formula, $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$ and $S_{n} = \frac{n}{2} (a + l)$ , give identical results — pick whichever needs the least extra work for the values already known.

Worked examples

Worked example

An arithmetic sequence begins $5, 9, 13, \dots$ .

Find

(a) the 20th term and the sum of the first 20 terms,

(b) the sum of the 5th to the 20th terms inclusive.

Show worked solution

Here $a = 5$ and $d = 4$ . (a):

u_{20} = 5 + 19 (4) = 5 + 76 = 81

and:

S_{20} = \frac{20}{2} (2 (5) + 19 (4)) = 10 (10 + 76) = 860

Check with the other form:

S_{20} = \frac{20}{2} (a + l) = 10 (5 + 81) = 860

(b) The 5th term is:

u_{5} = 5 + 4 (4) = 21

and the block from the 5th to the 20th contains $20 - 5 + 1 = 16$ terms, so its sum is:

\frac{16}{2} (21 + 81) = 8 (102) = 816

Check by subtraction:

S_{20} - S_{4} = 860 - \frac{4}{2} (2 (5) + 3 (4)) = 860 - 2 (22) = 860 - 44 = 816

Worked example

The 3rd term of an arithmetic series is $11$ and the 7th term is $27$ .

Find the first term and common difference, and hence the number of terms needed for the sum to reach $903$ .

Show worked solution

The 7th term exceeds the 3rd by four common differences:

4 d = 27 - 11 = 16

so $d = 4$ .

Then:

u_{3} = a + 2 d = 11

gives $a = 11 - 8 = 3$ .

The sum is:

S_{n} = \frac{n}{2} (2 (3) + (n - 1) 4) = \frac{n}{2} (4 n + 2) = n (2 n + 1) = 2 n^{2} + n

Setting $2 n^{2} + n = 903$ gives:

2 n^{2} + n - 903 = 0

with discriminant:

1^{2} - 4 (2) (- 903) = 1 + 7224 = 7225 = 8 5^{2}

so:

n = \frac{- 1 \pm 85}{4}

The positive root is:

n = \frac{84}{4} = 21

Check:

S_{21} = 21 (2 \times 21 + 1) = 21 \times 43 = 903

1.04

Geometric sequences and series

Geometric sequences 1.04

Definitions

Geometric sequence: one in which consecutive terms have a constant common ratio $r$ , so $\frac{u _{n + 1}}{u _{n}} = r$ for every $n$ .

Key results

$n$ th term: $u_{n} = a r^{n - 1}$ .
For three consecutive terms of a geometric sequence, the middle term satisfies $(u_{2})^{2} = u_{1} u_{3}$ — the standard test for whether three given quantities form a geometric progression.

Notes

A negative $r$ makes the terms alternate in sign; $∣ r ∣ < 1$ makes them shrink towards zero; $∣ r ∣ > 1$ makes them grow without bound.
To find the least $n$ for which $a r^{n - 1}$ passes some threshold, take logarithms of both sides — and reverse the inequality when dividing by $lo g r$ if $0 < r < 1$ , since $lo g r$ is then negative.
The test $(u_{2})^{2} = u_{1} u_{3}$ only confirms three numbers COULD be consecutive geometric terms — it says nothing about the value of $r$ itself, which still needs $\frac{u _{2}}{u _{1}}$ or $\frac{u _{3}}{u _{2}}$ to find.

Geometric series 1.04

Key results

Sum of the first $n$ terms: $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ for $r \neq = 1$ .
Multiplying numerator and denominator by $- 1$ gives the equivalent form $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ , which keeps the arithmetic positive when $r > 1$ .
If $r = 1$ every term equals $a$ , so the sum is simply $S_{n} = na$ and the standard formula does not apply.

Notes

The formula is derived by writing out $S_{n}$ , multiplying it by $r$ , and subtracting: almost every term cancels, leaving $S_{n} - r S_{n} = a - a r^{n}$ .
Always check whether $r$ is positive or negative before choosing between the two equivalent forms of the formula — the $\frac{a ( r ^{n} - 1 )}{r - 1}$ version avoids a leading minus sign when $r > 1$ , but both give the same value regardless.

Convergence and the sum to infinity 1.04

Definitions

Convergent geometric series: one whose partial sums $S_{n}$ approach a finite limit as $n$ increases, which happens exactly when $∣ r ∣ < 1$ .

Key results

Sum to infinity: $S_{\infty} = \frac{a}{1 - r}$ , valid only for $∣ r ∣ < 1$ .
This follows from $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ , because $r^{n} \to 0$ as $n \to \infty$ whenever $∣ r ∣ < 1$ .

Notes

The condition $∣ r ∣ < 1$ is essential and easy to forget: a geometric series with $∣ r ∣ \geq 1$ does not converge, and $S_{\infty}$ is simply undefined in that case, however tempting it is to substitute into the formula regardless.
If a question gives $S_{\infty}$ and one of $a$ or $r$ , use $a = S_{\infty} (1 - r)$ to recover the other; always confirm the resulting $r$ satisfies $∣ r ∣ < 1$ .
$S_{\infty}$ is a limit, not a term that is ever actually reached — no finite partial sum $S_{n}$ equals $S_{\infty}$ exactly, it only gets arbitrarily close as $n$ grows.

Worked examples

Worked example

A geometric series has first term $8$ and common ratio $\frac{1}{2}$ .

(a) Find the sum to infinity.

(b) Find the least value of $n$ for which $S_{n}$ exceeds $15.9$ .

Show worked solution

(a) Since:

∣ r ∣ = \frac{1}{2} < 1

the series converges, so:

S_{\infty} = \frac{a}{1 - r} = \frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16

(b):

S_{n} = \frac{8 ( 1 - ( \frac{1}{2} ) ^{n} )}{1 - \frac{1}{2}} = 16 (1 - 2^{- n})

Requiring:

16 (1 - 2^{- n}) > 15.9

gives:

1 - 2^{- n} > 0.99375

so:

2^{- n} < 0.00625

i.e.

$2^{n} > 160$ .

Since $2^{7} = 128$ and $2^{8} = 256$ , the least value is $n = 8$ .

Check:

S_{8} = 16 (1 - \frac{1}{256}) = 15.9375 > 15.9

while:

S_{7} = 16 (1 - \frac{1}{128}) = 15.875 < 15.9

Worked example

The 1st, 5th and 13th terms of an arithmetic sequence with non-zero common difference $d$ are also, in that order, the first three terms of a geometric sequence.

Show that $a = 4 d$ and find the common ratio.

Show worked solution

The three terms are $a$ , $a + 4 d$ and $a + 12 d$ .

For a geometric sequence the middle term satisfies:

(a + 4 d)^{2} = a (a + 12 d)

Expanding:

a^{2} + 8 a d + 16 d^{2} = a^{2} + 12 a d

so $16 d^{2} = 4 a d$ , and since $d \neq = 0$ this gives $4 d = a$ .

The three terms are therefore $4 d$ , $8 d$ and $16 d$ , so the common ratio is:

r = \frac{8 d}{4 d} = 2

confirmed by:

\frac{16 d}{8 d} = 2

1.04

Sigma notation

Sigma notation 1.04

Definitions

Sigma notation: $i = 1 \sum n f (i)$ means the sum of $f (i)$ as the index $i$ runs through the integers from $1$ to $n$ inclusive.

Key results

The number of terms in $i = p \sum q$ is $q - p + 1$ .
$i = 1 \sum n (3 i + 2)$ is an arithmetic series with $a = 5$ and $d = 3$ ; $i = 1 \sum n 4 (\frac{1}{2})^{i - 1}$ is a geometric series with $a = 4$ and $r = \frac{1}{2}$ .

Notes

Both the binomial expansion and the arithmetic and geometric series formulas can be written in this notation, and recognising the correspondence between a sigma expression and the sequence it generates is itself a common exam skill.
To identify the type of series, write out the first two or three terms explicitly: a constant difference means arithmetic, a constant ratio means geometric.
Changing the starting index (e.g. from $i = 1$ to $i = 0$ ) shifts every term by one position — always re-derive $a$ , $d$ or $r$ from the actual first term written out, not from assuming the index always starts at the 'usual' first term.

1.04

The binomial expansion

Binomial coefficients and the expansion of $(1 + x)^{n}$ 1.04

Definitions

Binomial coefficient: $(k n) = \frac{n !}{k ! ( n - k )!}$ , also written $^{n} C_{k}$ , counting the number of ways of choosing $k$ objects from $n$ .

Key results

For a positive integer $n$ : $(1 + x)^{n} = k = 0 \sum n (k n) x^{k}$ , so the coefficient of $x^{k}$ is exactly $(k n)$ , counting the ways of choosing which $k$ of the $n$ factors contribute the $x$ term.
$(0 n) = (n n) = 1$ and $(1 n) = (n - 1 n) = n$ .
The coefficients are symmetric, $(k n) = (n - k n)$ , and are the rows of Pascal's triangle.

Notes

The expansion of $(1 + x)^{n}$ for positive integer $n$ terminates after $n + 1$ terms — it is a finite, exact identity, not an approximation.
$(k n)$ can also be read directly off Pascal's triangle (row $n$ , entry $k$ , both counting from $0$ ) — useful as a quick check for small $n$ without evaluating the factorial formula.

Expanding $(a + b)^{n}$ and finding a particular term 1.04

Key results

General term of $(a + b)^{n}$ : $(k n) a^{n - k} b^{k}$ for $k = 0, 1, \dots, n$ , so the powers of $a$ fall as the powers of $b$ rise, and the two exponents always sum to $n$ .

Method

Identify $a$ , $b$ and $n$ from the bracket exactly as written, including any coefficient or sign.
Write the general term $(k n) a^{n - k} b^{k}$ without yet simplifying.
Set the power of $x$ in that term equal to the power you need, and solve for $k$ .
Substitute that $k$ back and evaluate, applying the power to the whole of $b$ including its coefficient and sign.

In practiceFind the coefficient of $x^{3}$ in $(2 - 3 x)^{7}$ .

$(k 7) 2^{7 - k} (- 3 x)^{k}$ general term, with $a = 2$ , $b = - 3 x$ , $n = 7$
$k = 3$ the power of $x$ is $k$
$(3 7) 2^{4} (- 3)^{3} = 35 \times 16 \times (- 27) = - 15120$ the cube applies to the $- 3$ too

Notes

Raising the whole of $b$ to the power $k$ is where most marks are lost: in $(2 - 3 x)^{6}$ the $k = 3$ term uses $(- 3 x)^{3} = - 27 x^{3}$ , not $- 3 x^{3}$ .
For an expansion in ascending powers of $x$ , list $k = 0, 1, 2, \dots$ in order; for descending powers, start from $k = n$ .
When the two exponents in the general term must sum to $n$ , use that as a check: after solving for $k$ , confirm $(n - k) + k = n$ still holds before evaluating the term.

Worked examples

Worked example

Find the first four terms, in ascending powers of $x$ , of the binomial expansion of $(1 + 2 x)^{5}$ .

Show worked solution

Using $(k 5) (2 x)^{k}$ for $k = 0, 1, 2, 3$ : the term for $k = 0$ is $(0 5) = 1$ .

For $k = 1$ :

(1 5) (2 x) = 5 (2 x) = 10 x

For $k = 2$ :

(2 5) (2 x)^{2} = 10 \times 4 x^{2} = 40 x^{2}

For $k = 3$ :

(3 5) (2 x)^{3} = 10 \times 8 x^{3} = 80 x^{3}

So:

(1 + 2 x)^{5} = 1 + 10 x + 40 x^{2} + 80 x^{3} + \dots

Worked example

Find the coefficient of $x^{3}$ in the expansion of $(2 - 3 x)^{6}$ .

Show worked solution

The general term is:

(k 6) 2^{6 - k} (- 3 x)^{k}

and $x^{3}$ requires $k = 3$ .

Then:

(3 6) = \frac{6 !}{3 ! 3 !} = 20

$2^{6 - 3} = 2^{3} = 8$ , and:

(- 3 x)^{3} = - 27 x^{3}

Multiplying:

20 \times 8 \times (- 27) = - 4320

so the coefficient of $x^{3}$ is $- 4320$ .

Note the sign and the cube of the whole of $- 3 x$ : using $- 3 x^{3}$ instead would give $- 480$ , which is wrong.

Sequences and series

Sequences and series: the language 1.04

Arithmetic sequences and series

Arithmetic sequences 1.04

Arithmetic series 1.04

Geometric sequences and series

Geometric sequences 1.04

Geometric series 1.04

Convergence and the sum to infinity 1.04

Sigma notation

Sigma notation 1.04

The binomial expansion

Binomial coefficients and the expansion of (1+x)n 1.04

Expanding (a+b)n and finding a particular term 1.04

Binomial coefficients and the expansion of $(1 + x)^{n}$ 1.04

Expanding $(a + b)^{n}$ and finding a particular term 1.04