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1.6 Series

Syllabus
9709–2028–2029
Topic
1.6
Level
AS

Learning objectives

1.6.1Binomial expansion• use the expansion of (a + b)n, where n is a positive integer Including the notations n rfp and n!. Knowledge of the greatest term and properties of the coefficients are not required.1.6.2Progressions• recognise arithmetic and geometric progressions1.6.3Progression formulae• use the formulae for the nth term and for the sum of the first n terms to solve problems involving arithmetic or geometric progressions Including knowledge that numbers a, b, c are 'in arithmetic progression' if 2b = a + c (or equivalent) and are 'in geometric progression' if b2 = ac (or equivalent). Questions may involve more than one progression.1.6.4Geometric convergence• use the condition for the convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progression.

Use the general binomial term to target one coefficient

For positive integer $n$:(a+b)^n=\sum_{r=0}^{n}inom nr a^{n-r}b^r,\qquad inom nr= rac{n!}{r!(n-r)!}.

Write the general term T_{r+1}=inom nr a^{n-r}b^r. If a variable occurs in aa or bb, equate its resulting power to the requested power, solve for the integer rr, then simplify the entire term.

In $(1+2x)^5$, the $x^2$ term has $r=2$:inom52(2x)^2=10\cdot4x^2=40x^2.

If the second term is negative, include its sign inside the power. For example, the sign of (−3x)r(-3x)^r depends on whether rr is odd or even.

inom nr is only the combinatorial factor, not the whole expansion term. Greatest-term methods and special coefficient properties are not required here.

Classify a progression by its constant adjacent change

Type Adjacent check Structural rule
Arithmetic progression (AP) uk+1−uk=du_{k+1}-u_k=d is constant add the same dd each step
Geometric progression (GP) uk+1/uk=ru_{k+1}/u_k=r is constant where defined multiply by the same rr each step

Calculate at least two consecutive differences or ratios. Classify only if the same value continues across all given adjacent pairs; then use that value as dd or rr.

11,7,3,−1,…11,7,3,-1,\ldots is AP with d=−4d=-4. 3,−6,12,−24,…3,-6,12,-24,\ldots is GP with r=−2r=-2; its signs alternate because the ratio is negative.

A pattern that merely rises, falls or alternates need not be AP or GP. Do not infer a common ratio from non-consecutive terms, and do not divide by a zero term.

Translate finite progression conditions into equations

Progression nnth term First nn terms
AP un=a+(n−1)du_n=a+(n-1)d S_n= rac n2[2a+(n-1)d]
GP un=arn−1u_n=ar^{n-1} S_n= rac{a(1-r^n)}{1-r} for $r
e1$

Three numbers $a,b,c$ are in AP when $2b=a+c$; they are in GP when $b^2=ac$ (with the stated order and real-number/domain conditions).

Define the first term and difference/ratio for each progression. Turn every stated term, sum or three-term relationship into an equation, solve the simultaneous system, reject values that violate the original order or denominator conditions, and substitute back.

For the AP $5,8,11,\ldots$:u_n=5+3(n-1)=3n+2,\qquad S_n= rac n2[10+3(n-1)]= rac{n(3n+7)}2.

unu_n is one term; SnS_n is the sum of the first nn terms. A problem may link more than one progression, so do not assume they share the same first term, dd or rr unless stated.

An infinite geometric series converges only when its ratio has magnitude below one

For first term a and common ratio r, S∞=a/(1−r) exists only if |r|<1. The partial sums approach a finite limit because later terms shrink to zero.

Check convergence before using the formula. A negative r gives alternating partial sums, but still converges when |r|<1.

3−1.5+0.75−… has a=3,r=−0.5 and S∞=3/1.5=2; the alternating signs do not prevent convergence.

A ratio close to 1 may converge slowly, while r=1 or −1 does not produce a finite infinite sum.