01 / Decode the notation

Make the
pattern click.

An interactive guide to sequences, sigma notation and series — including both parts of the worked examples.

Edexcel A level Mathematics · 9MA0 Pure

One rule. Many terms.

1 + ½ + ¼ + ⅛ + ⋯

Infinitely many terms. A finite sum.

What this coversThe sequences and series skills behind your screenshot: sigma, arithmetic and geometric methods, convergence, logarithmic cancellation, proofs, recurrence relations and modelling. It is a focused topic guide, not the whole A level course.

A sequence is a list

3, 6, 12, 24, …

u_n means the nth term. Here u_n = 3 × 2ⁿ⁻¹.

A series adds the list

3 + 6 + 12 + 24 + …

S_N means the sum of the first N terms. Here S₄ = 45.

Read sigma by substitution

Move the limits

Σ says “add”. Use every integer from the lower limit to the upper limit, including both ends.

First index m Last index M

Term rule

The index letter is a label: k, r, i or n can do the same job. In the worked examples, r is the index; the common ratio is ½. This guide uses q for a ratio to keep them distinct.

Three rules that prevent most mistakes

Number of terms = M − m + 1For integer limits m ≤ M. From 4 to 10 there are 7 terms.

∑(A_k + B_k) = ∑A_k + ∑B_kThe sums must use the same limits. A constant factor can also come outside Σ.

∑_k=m^M u_k = S_M − S_m−1If S_j = u₁ + ⋯ + u_j, with S₀ = 0.

Also: ∑_k=1^N 1 = N and ∑_k=1^N k = N(N + 1)/2. A constant is added once per term: ∑_k=3⁷ 5 = 25.

02 / The same difference

Arithmetic sequences.

Add the same amount each time. That amount, d, can be positive, zero or negative.

Build a sequence

Drag to explore

First term a Common difference d Number of terms N

First term

Nth term u_N

Sum S_N

Find a term

u_n = a + (n − 1)d

The first term has had zero additions. Reaching the nth term takes n − 1 additions.

If u₃ = 11 and u₈ = 31, subtract: 5d = 20, so d = 4. Then a + 2(4) = 11, giving a = 3.

Add the terms

S_N = N2(2a + (N − 1)d)

= N2(a + l)

l is the last term. The sum is the number of terms × the average of the first and last terms.

Proof you should know: reverse and add

S_N = a + (a + d) + ⋯ + l
S_N = l + (l − d) + ⋯ + a
2S_N = N(a + l)

Each vertical pair has the same total a + l, and there are N pairs. Divide by 2, then use l = a + (N − 1)d.

Derive the sum of the first N natural numbers

1 + 2 + ⋯ + N = N(1 + N)/2.

This is the arithmetic sum with a = 1 and d = 1.

Shifted sigma example

∑_k=3⁷ (2k + 1) = 7 + 9 + 11 + 13 + 15

There are 7 − 3 + 1 = 5 terms. So the sum is 5(7 + 15)/2 = 55. The first included term is 7.

03 / The same multiplier

Geometric sequences.

Multiply by the same number, q, each time. Edexcel usually writes the common ratio as r.

From terms to a sum

First term a Common ratio q Number of terms N

Ratio q

Nth term u_N

Sum S_N

Nth term

u_n = aqⁿ⁻¹

q = next term ÷ current term, whenever the current term is non-zero. Check more than one pair.

If u₃ = 12 and u₆ = 96, divide: q³ = 96/12 = 8, so q = 2. Then a = 12/2² = 3.

An even power can give two signs: q² = 4 means q = ±2 unless other information rules one out.

Finite geometric sum

S_N = a(1 − q^N)1 − q q ≠ 1

= a(q^N − 1)q − 1

Both forms work for q < 1 or q > 1. Match the order of subtraction in the numerator and denominator.

If q = 1: every term is a, so S_N = Na.

Proof you should know: multiply and subtract

S_N = a + aq + aq² + ⋯ + aq^N−1
qS_N = aq + aq² + ⋯ + aq^N
S_N − qS_N = a − aq^N
(1 − q)S_N = a(1 − q^N)

The middle terms cancel. Divide by 1 − q, provided q ≠ 1.

How sigma changes the first term

∑_k=m^M Cq^k = Cq^m(1 − q^M−m+1)1 − q

First included term a = Cq^m; number of terms N = M − m + 1. Use the ordinary sum formula with these values. If q = 1, the sum is C(M − m + 1).

For k = 4 to 7, ∑20(½)^k has 4 terms, starts at 5/4, and sums to 75/32.

Write a series using sigma

5 + 15 + 45 + 135 + 405
= ∑_k=1⁵ 5 × 3^k−1 = ∑_j=0⁴ 5 × 3^j

Both are valid. Substitute the first, second and last index values to check.

04 / A limit, not a last term

Can infinity add up?

Move the ratio to see partial sums approach a limit, oscillate, or grow without settling.

The infinity lab

Worked examples loaded

Coefficient C Ratio q Start index m Terms drawn N

Partial sumLimit, when it exists

First included term a = Cq^m

Sum of N drawn terms

Sum to infinity

The condition comes first

S_∞ = a1 − q For a non-zero geometric series, only when |q| < 1.

|q| < 1 means −1 < q < 1. Negative ratios can converge: the terms alternate in sign while shrinking.

As N → ∞, q^N → 0. Hence a(1 − q^N)/(1 − q) → a/(1 − q).

How much is left?

R_N = S_∞ − S_N = aq^N1 − q

This is the signed remainder after N terms, when |q| < 1. The absolute error is |R_N|.

The remaining tail starts at aq^N. For the worked examples, the infinite tail starts at r = 4, so a = 20(½)⁴.

Know the boundary cases

Ratio q	Non-zero terms	Sum to infinity?
0 < q < 1	Shrink with the same sign	Yes
−1 < q < 0	Alternate and shrink	Yes
q = 1	Stay constant	No
q = −1	Alternate without shrinking	No: partial sums alternate
\|q\| > 1	Grow in magnitude	No

q = 0 gives a, 0, 0, …, so the sum is a. If every included term is zero, the sum is 0 regardless of the multiplier. Terms tending to zero is necessary for a convergent series, but is not sufficient for general series.

05 / Let the middle disappear

Telescoping logarithms.

Add the logs, multiply the fractions, then cancel common factors. This is the idea in part (ii).

Last index N Shift c Log base b

Reverse every fraction

For long sums the middle factors are shortened with “…”. All displayed crossed-out factors cancel with matching factors in the full product.

The three log laws

log_b A + log_b B = log_b(AB)
log_b A − log_b B = log_b(A/B)
p log_b A = log_b(A^p)

For real logs, b > 0, b ≠ 1 and A, B > 0. Also log_b(b^p) = p and log_b 1 = 0.

log(A + B) is generally not log A + log B. The product law needs a common base.

A second way: cancel differences

Use log(A/B) = log A − log B:

(log₅ 3 − log₅ 2)
+ (log₅ 4 − log₅ 3)
+ ⋯ + (log₅ 50 − log₅ 49)

Everything in the middle cancels, leaving log₅ 50 − log₅ 2 = log₅ 25 = 2.

Generalise the pattern

∑_n=L^U log_bn + c + 1n + c = log_bU + c + 1L + c

For integer L ≤ U, L + c > 0, b > 0 and b ≠ 1. Write out boundary terms if the shift differs from 1: a larger gap can leave several boundary factors.

Telescoping without logarithms

∑_n=1^N 1n(n + 1) = ∑_n=1^N (1/n − 1/(n + 1))
= 1 − 1/(N + 1)

The same cancellation idea works after splitting into partial fractions. As N → ∞, this sum tends to 1. This is an extension connecting the pattern to algebra.

06 / Your exact question

One step at a time.

Choose a part and use Next step. Try saying why each line follows before revealing it.

Reveal the full written solution

How to write a clear exam answer

Translate the limits. Show the first included term or the first and final fractions.
Name the method. Identify a geometric series or use a log law.
Show the reason. State |q| < 1 for an infinite geometric sum, or show the cancellation in a telescoping product.
Give an exact result. Here the answers are 5/2 and 2. A decimal approximation alone is weak evidence for a “show that” question.

In (i), a = 20(½)⁴ = 5/4. In (ii), n = 48 gives 50/49. These are the two boundary substitutions to check.

07 / Finish the topic

The exam toolkit.

Choose a method, solve for unknowns, and connect sequences to models and recurrence relations.

Start by asking: what changes?

Clue	Method	First check
Same difference	Arithmetic nth term / sum	Find d and the first included term
Same ratio	Geometric nth term / finite sum	Count terms; q = 1 is separate
∞ on a geometric sum	Sum to infinity	Check \|q\| < 1
Sum of logs of neighbouring fractions	Product law, then telescope	Common base; positive arguments
Unknown in an exponent	Isolate the power; take logs	Positive log arguments; integer n
Next term given from previous term	Iterate the recurrence	Use the stated starting value

Use logs to find the number of terms

For 100 + 80 + 64 + ⋯, find the smallest number of terms whose sum exceeds 490.

S_N = 100(1 − 0.8^N)/(1 − 0.8)
500(1 − 0.8^N) > 490
0.8^N < 0.02
N ln(0.8) < ln(0.02)
N > ln(0.02)/ln(0.8) ≈ 17.53

So the smallest integer is N = 18.

The inequality reversesln(0.8) is negative. Dividing by a negative number reverses < to >. Check the neighbouring integers: S₁₇ ≈ 488.74 and S₁₈ ≈ 490.99.

For equations q^N = t with q > 0, q ≠ 1 and t > 0: N = ln(t)/ln(q). With negative q, handle the sign and parity separately; do not take a real log of a negative number.

Precision question: when is the error small enough?

If a = 10 and q = ½, then |R_N| = 20(½)^N. To make the error less than 0.01:

(½)^N < 0.0005 ⇒ N > 10.965…

The smallest integer is 11. Always interpret a strict inequality correctly; if the threshold were exactly an integer, “greater than” would require the next integer.

Recurrence relations

A recurrence tells you how to generate the next term. You also need a starting value.

u₁ = 3, u_n+1 = 2u_n
3, 6, 12, 24, …

u_n+1 = u_n + d is arithmetic. u_n+1 = qu_n is geometric.

Try a recurrence that is neither

u₁ = 1, u_n+1 = 2u_n + 1

Terms: 1, 3, 7, 15, … . Their differences and ratios are not constant. Here u_n = 2ⁿ − 1; substitute to check it fits the recurrence and start.

Increasing, decreasing, periodic

Increasing: u_n+1 > u_n for every allowed n. Example: 2n.

Decreasing: u_n+1 < u_n. Example: 3 + 1/n for n ≥ 1.

Periodic: the sequence repeats. u₁ = 3, u_n+1 = 1/u_n gives 3, ⅓, 3, ⅓, …, with period 2.

To prove increasing or decreasing, show the sign of u_n+1 − u_n for all allowed n. A few terms suggest a pattern; they do not prove it.

Modelling: put time on the terms

Same percentage

A £2000 item loses 10% of its value per year. After t complete years, V_t = 2000(0.9)^t, with V₀ = 2000.

A 5% increase uses 1.05; a 5% decrease uses 0.95. Repeated percentage change is geometric.

Regular payments

Pay £100 at each year-end into an account earning 2% annually. Just after payment n:

V_n+1 = 1.02V_n + 100, V₀ = 0.

Thus V_n = 100[1 + 1.02 + ⋯ + 1.02ⁿ⁻¹] = 5000(1.02ⁿ − 1). The balances themselves do not have a constant ratio.

State the assumptions: fixed rates, payment timing, and no extra charges. Check whether the first time is t = 0 or t = 1. Round money at the end and judge whether the model stays realistic.

Mistakes to catch before moving on

Tempting mistake	Correct habit
Use C as the first term of ∑_k=m Cq^k	Substitute k = m: first term is Cq^m
Use M − m terms	Include both ends: M − m + 1
Use aqⁿ when u₁ = a	Use aqⁿ⁻¹; u₁ needs q⁰
Claim convergence because q < 1	Need −1 < q < 1; q = −2 fails
Cancel terms in a sum of fractions	Rewrite as a product or differences first
Round the answer for n to the nearest integer	Use the inequality and check neighbouring integers
Prove a “show that” using only decimals	Show the algebra and keep an exact answer

08 / Make it stick

Try it yourself.

Original practice, from boundary substitutions to exam-style reasoning. Write your working, then check the answer.

Ready for this topic?

Use these checks to decide what to revisit.

I can read a shifted sigma sum and count its terms.I can choose arithmetic or geometric formulas and prove both sum formulas.I can find the first included term and justify convergence using |q| < 1.I can telescope a log sum and explain exactly what cancels.I can use logs to solve for n and handle integer inequalities.I can generate recurrence terms and interpret a model's timing.

If one feels uncertain, return to its lesson, change the numbers and explain the method aloud.

Related reference: Pearson Edexcel 9MA0 specification, sections 4.2–4.6, with log laws used for the worked examples. Original explanations and practice; not an official Pearson resource. Log telescoping is an application of those skills.

Make thepattern click.