Sequences

Term-to-Term Rules #

What gets you from each term to the next.

A sequence has a rule that gets you from each term to the next one

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3, 7, 11, 15. The gaps are all the same size, so the rule is add 4.

2, 4, 8, 16, 32 does not add. Each term doubles, so the gaps stretch.

The nth Term #

The gap is the times table it is built on.

The common difference is the times table the sequence is built on

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3, 7, 11, 15. Every step adds the same 4 dots.

The 4 times table grows the same way, one dot ahead of ours each time.

So the rule is 4n − 1, and against position the terms lie on a line.

Sequences Worth Knowing #

Squares, triangle numbers and Fibonacci.

Some sequences are worth recognizing on sight rather than working out

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1, 4, 9, 16 are the square numbers, each one a square you can draw.

1, 3, 6, 10 are the triangle numbers: add 2, then 3, then 4.

Fibonacci adds the two before it: 1, 1, 2, 3, 5, 8, 13.

Quadratic Sequences #

When the gaps between the gaps are steady.

When the gaps themselves change by a fixed amount the rule involves n squared

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3, 8, 15, 24 climbs faster and faster, so the plot bends.

The gaps go 5, 7, 9. Their gaps are a steady 2, which is the signal.

Half of the steady 2 is 1, so try $n^2$n². Take $n^2$n² off and 2, 4, 6, 8 is what is left.

What is left climbs by a steady 2, so it is 2n — and the rule is $n^2+2n$n² + 2n.

Geometric Sequences #

Multiplying by the same amount each time.

A geometric sequence multiplies by the same amount every time instead of adding

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3, 6, 12, 24 doubles each step. The curve gets steeper without limit.

Divide any term by the one before and you always get the same 2 — the common ratio.

Term n stands n − 1 doublings past the start, so the nth term is $3\times 2^{n-1}$3 × 2ⁿ⁻¹.

A multiplier under 1 shrinks instead, closing in on zero without arriving.

Geometric Sequences with a Surd Ratio #

Multiply by $\sqrt{2}$√2 twice and you have multiplied by 2.

A common ratio can be a surd; multiplying by it twice multiplies by a whole number

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A common ratio can be a surd: each term here is $\sqrt{2}$√2 times the one before.

Multiplying by $\sqrt{2}$√2 twice multiplies by 2, so every second term is a whole number.

The nth term is $a\times r^{n-1}$a × rⁿ⁻¹ as usual. An even power of $\sqrt{2}$√2 is a power of 2: ${\left(\sqrt{2}\right)}^{10}=2^5$(√2)¹⁰ = 2⁵.

For three terms in a row, the ratios match: $x/1=3/x$x / 1 = 3 / x, so $x^2=3$x² = 3 and $x=\sqrt{3}$x = √3.

Sigma Notation #

The stretched S that says: add these up.

The stretched Greek S is an instruction to add the terms its expression describes

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$\sum _1^{4}n$Σ n from 1 to 4 unpacks to 1 + 2 + 3 + 4 = 10.

The expression under it names each term: 2n at n = 1, 2, 3, 4 gives 2, 4, 6, 8.

The little numbers are the limits: start at n = 1, stop at 4. Four terms, no more.

Raise the top limit and the sum grows: to 5, the same $\sum$Σ adds one more term, the 5.

The Sum of an Arithmetic Series #

Fold the ends in; every pair matches.

Fold an arithmetic series end to end and every pair adds to the same first-plus-last

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Fold the list in half: 1 with 10, 2 with 9, 3 with 8 — every pair makes the same 11.

Ten terms fold into five pairs, so the sum is 5 × 11 = 55. No slog required.

Any arithmetic series folds the same way: $\frac{n}{2}$n/2 pairs, each worth first plus last.

Ten terms from 2 up to 29: five pairs, each worth 31, so the sum is 155.

The Sum of a Geometric Series #

Double it, subtract it, watch it collapse.

Doubling a geometric series and subtracting it collapses the sum to two terms

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Double the sum and subtract the original: the overlapping terms wipe each other out.

What survives is tiny: S = 16 − 1 = 15 — the next term, minus the first term.

In letters the same collapse gives the formula: $S=\frac{a\left(r^n-1\right)}{r-1}$S = a(rⁿ − 1)/(r − 1).

Check it by hand: 3 + 6 + 12 + 24 = 45, and the formula says 3 × 15 = 45. Agreed.

The nth Term from the Sum #

Getting a single term back from a formula for the sum.

The sum of n terms minus the sum of n − 1 terms leaves the nth term: $u_n=S_n-S_{n-1}$uₙ = Sₙ − Sₙ₋₁

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$S_4$S₄ and $S_3$S₃ share their first three terms, so $S_4-S_3$S₄ − S₃ leaves $u_4$u₄. In general, $u_n=S_n-S_{n-1}$uₙ = Sₙ − Sₙ₋₁.

Given $S_n=n^2+2n$Sₙ = n² + 2n, replace n with n − 1 to get $S_{n-1}=n^2-1$Sₙ₋₁ = n² − 1, then subtract: $u_n=2n+1$uₙ = 2n + 1.

$S_n-S_{n-1}$Sₙ − Sₙ₋₁ needs $n\ge 2$n ≥ 2, so $u_1$u₁ comes from $S_1$S₁. With $S_n=n^2+2n+1$Sₙ = n² + 2n + 1, $u_1=4$u₁ = 4, while 2n + 1 would give 3.

The Sum to Infinity #

Endless terms, finite total.

A geometric series with a shrinking ratio settles on the finite value a over one minus r

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The halving story again: $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$1/2 + 1/4 + 1/8 + … fills the bar toward 1, never past it.

The condition is a shrinking multiplier: r strictly between −1 and 1. At r = 2 the sum diverges.

In the finite sum, $r^n$rⁿ withers to nothing as n grows — what is left is $\frac{a}{1-r}$a/(1 − r).

8 + 4 + 2 + 1 + … settles on $\frac{8}{1-\frac{1}{2}}=16$8/(1 − 1/2) = 16 — twice the first term, forever filled.

Recurrence Relations #

A rule and a place to stand.

A recurrence names the term-to-term rule in symbols, and it cannot start without a first term

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The term-to-term rule "add 4", written properly: each new u is the one before plus 4.

The rule alone floats free: until $u_1$u₁ is given, it names no sequence at all.

Fibonacci, from the sequences worth knowing, is a recurrence: add the two before.

Run it: from $u_1=2$u₁ = 2, doubling and subtracting 1 walks out 2, 3, 5, 9.

Proof by Induction: Summing 1 to n #

Topple the first domino, then every next one.

Proof by induction: true for one case, and each case forcing the next, covers all

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Knock one domino over, and make sure each one topples the next.

The claim: $1+2+\cdots +n=\frac{n\left(n+1\right)}{2}$1 + 2 + … + n = n(n + 1)/2. It holds at n = 1, so assume it at n.

Add (n + 1) to both sides and it tidies to $\frac{\left(n+1\right)\left(n+2\right)}{2}$(n + 1)(n + 2)/2 — the very next case.

True for 1 forces 2, which forces 3 — that chain is induction, and it never stops.

Proof by Induction: Divisibility #

The jump from one case to the next is a multiple.

Induction proves a divisibility claim by showing each jump adds another multiple

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The claim is about every n, and at n = 1 it is a number you can simply check.

Assume it at some k. The job is to force the very next case out of that assumption.

Subtract one case from the next. The constants cancel and the jump is 8 times a power.

So the next case is the assumed multiple of 4 plus another one, which is again a multiple.

Base case and step together carry the claim along the whole line, exactly as for a sum.