P2.4 - Sequences and series

Learning objectives

P2.4.1Sequences,Sequences, including those given by a formula for the nth term and those generated by a simple relation of the form x = f(x). n + 1 n.P2.4.2Arithmetic sequences and seriesUnderstand and work with The proof of the sum formula should be known. arithmetic sequences and series, including the formula for the nth term and the sum of a finite Σ Understanding of notation will expected. arithmetic series.; the sum of the first n natural numbers.P2.4.3Increasing sequences, decreasing sequences and periodicIncreasing sequences, decreasing sequences and periodic sequences.P2.4.4Geometric sequences and seriesUnderstand and work with For example, given the sum of a series students should be geometric sequences and series, able to use logs to find the value of n. including the formulae for the nth The proof of the sum formula for a finite series should be term and the sum of a finite known. geometric series.; the sum to infinity of a convergent geometric series, The sum to infinity may be expressed as S ∞. including the use of |r| < 1.P2.4.5Binomial expansion for positive integer powersExpand (a + bx)^n for positive integer n and use the notations n!, binomial coefficients and nCr.

A sequence is an ordered list whose position matters. A rule can give a term directly from its position, or generate each term from the previous one.

Rule type	What must be given	How a term is found
Explicit: $u_n=f(n)$	the formula and term number $n$	substitute $n$ directly
Recursive: $x_{n+1}=f(x_n)$	the recurrence and a starting value	calculate earlier terms in order

For $u_n=3n-1$ , the sequence begins $2,5,8,11,\ldots$ and $u_7=3(7)-1=20$ without finding terms 5 or 6.

For $x_1=2$ and $x_{n+1}=x_n^2-1$ , $x_2=3,\qquad x_3=8,\qquad x_4=63.$ Each input is the preceding term, so the starting value is part of the definition.

Keep the indices aligned: $x_{n+1}$ is the next term produced from $x_n$ . Do not replace $x_n$ by $n$ , or apply a recurrence without its required starting value.

An arithmetic sequence has a constant difference $d$ . Starting from first term $a$ , moving to term $n$ adds $d$ exactly $n-1$ times.

$u_n=a+(n-1)d,\qquad S_n=\frac n2\bigl(2a+(n-1)d\bigr)=\frac n2(a+l)$

The sum formula follows by writing the same finite sum forwards and backwards: $S_n=a+(a+d)+\cdots+[a+(n-1)d],$ $S_n=[a+(n-1)d]+\cdots+(a+d)+a.$ Each of the $n$ paired columns totals $2a+(n-1)d$ , so doubling and then halving gives the formula.

Sigma notation compresses an addition: $\sum_{r=1}^{n}u_r$ means $u_1+u_2+\cdots+u_n$ . Setting $a=1,d=1$ gives $1+2+\cdots+n=\frac{n(n+1)}2.$

For $7,11,15,\ldots$ , $a=7$ and $d=4$ . Thus $u_{20}=83$ and $S_{20}=\frac{20}{2}(7+83)=900.$

Term $n$ uses $n-1$ differences. In a finite sum, identify the number of terms before substituting; it is not automatically the value of the final term.

Classify a sequence by comparing consecutive terms or by identifying an exact repeating cycle.

Type	Test
Strictly increasing	$u_{n+1}>u_n$ for every relevant $n$
Strictly decreasing	$u_{n+1}<u_n$ for every relevant $n$
Periodic with period $p$	$u_{n+p}=u_n$ ; the order is the smallest positive such $p$

For an explicit sequence, inspect $u_{n+1}-u_n$ : a value always positive proves increasing, while one always negative proves decreasing. For $u_n=n^2+1$ , $u_{n+1}-u_n=2n+1>0$ for positive integer $n$ , so the sequence is increasing.

The sequence $4,\frac34,-\frac13,4,\frac34,-\frac13,\ldots$ repeats every three terms, so it is periodic of order $3$ . With a one-step deterministic recurrence, returning to an earlier value makes the same following cycle repeat.

A few rising or falling terms do not prove the pattern continues. Equality also fails the strict tests; and a stated period is not the order if a smaller positive period works.

A geometric sequence multiplies by the same common ratio $r$ each time. With first term $a$ , its terms and finite sum are

$u_n=ar^{n-1},\qquad S_n=\frac{a(1-r^n)}{1-r}\quad(r\ne1)$

Write $S_n=a+ar+\cdots+ar^{n-1}$ and multiply by $r$ . Subtracting $rS_n$ from $S_n$ cancels the middle terms, leaving $(1-r)S_n=a-ar^n,$ which gives the finite-sum formula.

Only when $|r|<1$ does $r^n\to0$ , so only then does the geometric sum converge to $S_\infty=\frac{a}{1-r}.$ A negative ratio may converge while the terms alternate signs.

For $a=5,r=0.8$ , requiring $S_n>24$ gives $25(1-0.8^n)>24$ , hence $0.8^n<0.04$ . Therefore $n>\frac{\log(0.04)}{\log(0.8)}\approx14.43,$ so the smallest integer is $15$ .

Use $ar^{n-1}$ for term $n$ , not $ar^n$ . Never use $S_\infty$ unless $|r|<1$ ; when solving an inequality with $0<r<1$ , dividing by $\log r<0$ reverses its direction.

For a positive integer $n$ , the binomial theorem gives every term of $(a+bx)^n$ without repeated multiplication.

$(a+bx)^n=\sum_{r=0}^{n}\binom nr a^{n-r}(bx)^r,\qquad \binom nr=\frac{n!}{r!(n-r)!}$

Here $n!=n(n-1)\cdots2\cdot1$ and $0!=1$ . The index $r$ is the power of $x$ , so choosing $r=0,1,2,\ldots$ produces terms in ascending powers.

For $(2-3x)^4$ , $\begin{aligned}(2-3x)^4={}&2^4+\binom41 2^3(-3x)+\binom42 2^2(-3x)^2\\&+\binom43 2(-3x)^3+(-3x)^4\\={}&16-96x+216x^2-216x^3+81x^4.\end{aligned}$

The sign belongs inside $(bx)^r$ : odd powers keep a negative sign and even powers become positive. This formula here is restricted to positive integer $n$ and terminates after $n+1$ terms.

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P2.4 - Sequences and series

Learning objectives

Generate a sequence directly or recursively

Build and sum an arithmetic sequence

Recognise increasing, decreasing and periodic sequences

Control finite and infinite geometric series

Expand a positive integer power with binomial coefficients

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