Introduction: Unlocking the Patterns of Sequences and Series
Welcome to the exciting world of Sequences and Series! This chapter is all about spotting patterns and creating mathematical formulas to describe them. Don't worry if it seems theoretical—sequences and series are everywhere, from calculating compound interest on your savings account to modeling population growth or decay.
In this chapter, we will learn how to describe patterns using simple rules, find any term in a long list, and, most importantly, calculate the total sum of these terms quickly.
P1.5: Core Concepts: Sequences and How They Grow
1. Defining Sequences and Series
A Sequence is an ordered list of numbers. Think of it as a set of instructions for building a list, one number after the next.
Example: 2, 4, 6, 8, 10, ... (The instruction is: start at 2 and add 2 each time.)
A Series is the sum of the terms in a sequence.
Example: 2 + 4 + 6 + 8 + 10 (The series is the total of those terms.)
The \(n\)-th term of a sequence is usually denoted by \(u_n\). This formula lets you jump straight to any term in the sequence without listing all the previous ones.
Key Notation: The Summation Symbol (\(\sum\))
We use the Summation Notation, or Sigma Notation (\(\sum\)), to efficiently write down a series.
$$\sum_{n=1}^{5} u_n$$
This means: "Find the sum of the terms \(u_n\), starting when \(n=1\) and stopping when \(n=5\)."
2. Sequences defined by Recurrence Relations
Sometimes, a sequence isn't defined by a direct formula for \(u_n\), but by a rule that tells you how to get the next term from the previous term. This is called a Recurrence Relation.
The general form is:
$$x_{n+1} = f(x_n)$$
To use this, you need a starting term, usually \(x_1\).
Example: A sequence is defined by \(x_{n+1} = 2x_n - 1\), starting with \(x_1 = 3\).
- \(x_1 = 3\)
- \(x_2 = 2(x_1) - 1 = 2(3) - 1 = 5\)
- \(x_3 = 2(x_2) - 1 = 2(5) - 1 = 9\)
Finding the Limit of a Convergent Sequence
If a sequence generated by a recurrence relation converges, it means the terms approach a fixed value, \(L\), as \(n\) gets very large (\(n \to \infty\)). At this limit, the next term is the same as the previous term, so \(x_{n+1} \approx x_n \approx L\).
Step-by-step to find the limit \(L\):
- Replace \(x_{n+1}\) with \(L\).
- Replace \(x_n\) with \(L\).
- Solve the resulting equation \(L = f(L)\).
1. Set \(L = \sqrt{2L + 3}\)
2. Square both sides: \(L^2 = 2L + 3\)
3. Rearrange: \(L^2 - 2L - 3 = 0\)
4. Factorise: \((L - 3)(L + 1) = 0\). Possible limits are \(L=3\) or \(L=-1\).
Since the terms of the original sequence must be positive (because of the square root), the limit must be positive. Thus, \(L=3\).
- Sequence: Ordered List (\(u_n\)).
- Series: The Sum of the list (\(S_n\) or \(\sum\)).
- Recurrence Relation: \(x_{n+1} = f(x_n)\).
- Limit: Solve \(L = f(L)\) for convergent sequences.
P1.5: Arithmetic Series (AP)
3. Understanding Arithmetic Progressions
An Arithmetic Progression (AP) or Arithmetic Series is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference, \(d\).
Analogy: Think of an AP as a steady salary that increases by a fixed amount (the common difference \(d\)) every year.
The \(n\)-th Term of an AP
If the first term is \(a\), the terms are:
$$a, \quad a+d, \quad a+2d, \quad a+3d, \quad \dots$$
The formula for the \(n\)-th term, \(u_n\):
$$\mathbf{u_n = a + (n-1)d}$$
(Why \(n-1\)? Because the first term (\(n=1\)) uses 0 differences.)
The Sum of an AP (\(S_n\))
The sum of the first \(n\) terms is denoted \(S_n\). There are two useful formulas:
1. Using the common difference \(d\):
$$\mathbf{S_n = \frac{n}{2} [2a + (n-1)d]}$$
2. Using the last term, \(l\) (where \(l = u_n\)):
$$\mathbf{S_n = \frac{n}{2} (a + l)}$$
Did you know? The formula \(S_n = \frac{n}{2} (a + l)\) works because terms in an AP are symmetrically spaced. For the sequence 2, 4, 6, 8, 10, the sum of the first and last (2+10=12) is the same as the sum of the second and second-to-last (4+8=12). You just multiply this pair sum by half the number of terms (\(n/2\)).
Example: Applying AP Formulas
Find the 20th term and the sum of the first 20 terms of the sequence 5, 8, 11, 14, ...
Step 1: Identify \(a\) and \(d\).
\(a = 5\) (First term)
\(d = 8 - 5 = 3\) (Common difference)
\(n = 20\)
Step 2: Find the 20th term (\(u_{20}\)).
$$u_n = a + (n-1)d$$
$$u_{20} = 5 + (20-1)3 = 5 + (19)(3) = 5 + 57 = 62$$
Step 3: Find the Sum of the first 20 terms (\(S_{20}\)).
Using \(S_n = \frac{n}{2} (a + l)\) where \(l = 62\):
$$S_{20} = \frac{20}{2} (5 + 62) = 10 (67) = 670$$
P1.5: Geometric Series (GP)
4. Understanding Geometric Progressions
A Geometric Progression (GP) or Geometric Series is a sequence where the ratio between consecutive terms is constant. This constant ratio is called the common ratio, \(r\).
Analogy: Think of a GP as investing money where the value increases by a fixed percentage (multiplied by \(r\)) each year. The growth is fast!
The \(n\)-th Term of a GP
If the first term is \(a\), the terms are:
$$a, \quad ar, \quad ar^2, \quad ar^3, \quad \dots$$
The formula for the \(n\)-th term, \(u_n\):
$$\mathbf{u_n = ar^{n-1}}$$
The Sum of a Finite GP (\(S_n\))
The sum of the first \(n\) terms is:
$$\mathbf{S_n = \frac{a(1-r^n)}{1-r}}$$
(Note: While \(r^n - 1\) in the numerator is also mathematically valid if you use \(r-1\) in the denominator, the syllabus generally uses the form above.)
The Sum to Infinity (\(S_\infty\))
This is one of the coolest parts of GPs! If the common ratio \(r\) is small enough, the terms eventually shrink almost to zero. If this happens, the series converges, and we can find its Sum to Infinity, \(S_\infty\).
The condition for convergence is crucial: The magnitude of the common ratio must be less than 1.
$$\mathbf{|r| < 1} \quad \text{or} \quad \mathbf{-1 < r < 1}$$
If this condition is met, the sum to infinity is:
$$\mathbf{S_\infty = \frac{a}{1-r}}$$
Common Mistake to Avoid: Always check the convergence condition \(|r|<1\) before calculating \(S_\infty\). If \(|r| \ge 1\), the sum diverges (tends to infinity) and \(S_\infty\) cannot be found.
Example: Sum to Infinity
A geometric series has a first term \(a=10\) and a common ratio \(r=0.5\).
Step 1: Check convergence.
\(r = 0.5\). Since \(|0.5| < 1\), the series converges.
Step 2: Calculate \(S_\infty\).
$$S_\infty = \frac{a}{1-r} = \frac{10}{1 - 0.5} = \frac{10}{0.5} = 20$$
The sum of all infinitely many terms is exactly 20!
P1.5: The Binomial Expansion (Positive Integer Power)
5. Expanding \((1+x)^n\) when \(n\) is a Positive Integer
The Binomial Expansion gives us a quick way to expand brackets like \((x+y)^4\) without having to multiply everything out manually. In the AS unit (P1.5), we focus on the case where the power \(n\) is a positive whole number.
Pascal's Triangle and Binomial Coefficients
The coefficients (the numbers in front of the terms) come from Pascal's Triangle or can be calculated using Binomial Coefficients, written as \(\binom{n}{r}\) or \({}^nC_r\).
The formula for the Binomial Coefficient is:
$$\mathbf{\binom{n}{r} = \frac{n!}{r!(n-r)!}}$$
where \(n!\) (n factorial) means \(n \times (n-1) \times \dots \times 1\).
The Formula for \((1+x)^n\)
For a positive integer \(n\):
$$\mathbf{(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots + x^n}$$
Using binomial coefficients:
$$\mathbf{(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n}$$
Expanding \((a+b)^n\)
If the base terms are not 1 and \(x\), we can use the same pattern.
$$\mathbf{(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n}a^0b^n}$$
Tip: Notice that the powers of the first term (\(a\)) decrease from \(n\) to 0, and the powers of the second term (\(b\)) increase from 0 to \(n\). The sum of the powers in every single term must equal \(n\).
Remember the formula for the coefficients by observing the pattern:
$$\text{Term } 1: \frac{n(n-1)\dots}{r!}$$
The top term has \(r\) factors, and you divide by \(r!\).
P2.2: Advanced Series Applications
6. The Binomial Series (Rational Power)
In the A-Level syllabus (P2.2), we extend the binomial expansion to cases where the power \(n\) is any rational number (positive, negative, or a fraction).
When \(n\) is rational, the expansion produces an infinite series, unlike the finite expansion for positive integers.
The Formula for \((1+x)^n\) (for rational \(n\))
The formula is similar, but we use the "long" formula as the combination notation \(\binom{n}{r}\) only strictly applies when \(n\) is an integer.
$$\mathbf{(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots}$$
Validity Condition (\(|x|<1\))
Because this is an infinite series, it is only an accurate approximation (or exactly equal to the function) if it converges.
The expansion is valid only if:
$$\mathbf{|x| < 1}$$
This means \(-1 < x < 1\).
What if the bracket isn't \((1+x)\)?
If you need to expand \((a+bx)^n\), you must factor out \(a\) first to get it into the required \((1 + \text{something})^n\) format:
$$(a+bx)^n = \left[a\left(1 + \frac{b}{a}x\right)\right]^n = a^n \left(1 + \frac{b}{a}x\right)^n$$
In this case, the expansion is valid when the magnitude of the new 'x' term, \(\frac{b}{a}x\), is less than 1:
$$\left|\frac{b}{a}x\right| < 1$$
Example: Rational Power Expansion
Expand \(\frac{1}{(2-x)^2}\) up to the \(x^2\) term, and state the range of \(x\) for which the expansion is valid.
Step 1: Rewrite using negative power.
$$\frac{1}{(2-x)^2} = (2-x)^{-2}$$
Step 2: Factor out the constant to get the (1+...) form.
$$(2-x)^{-2} = \left[2\left(1 - \frac{x}{2}\right)\right]^{-2} = 2^{-2} \left(1 - \frac{x}{2}\right)^{-2} = \frac{1}{4} \left(1 - \frac{x}{2}\right)^{-2}$$
Step 3: Apply the binomial series formula.
Let \(N=-2\) and substitute \((-\frac{x}{2})\) for \(X\):
$$\left(1 + X\right)^N \approx 1 + NX + \frac{N(N-1)}{2!}X^2 + \dots$$
$$\left(1 - \frac{x}{2}\right)^{-2} \approx 1 + (-2)\left(-\frac{x}{2}\right) + \frac{(-2)(-3)}{2!}\left(-\frac{x}{2}\right)^2$$
$$\approx 1 + x + (3)\left(\frac{x^2}{4}\right)$$
$$\approx 1 + x + \frac{3}{4}x^2$$
Step 4: Multiply by the factorized constant.
$$(2-x)^{-2} = \frac{1}{4} \left(1 + x + \frac{3}{4}x^2 + \dots\right) = \mathbf{\frac{1}{4} + \frac{1}{4}x + \frac{3}{16}x^2 + \dots}$$
Step 5: Find the validity range.
The expansion is valid when \(\left|-\frac{x}{2}\right| < 1\).
This simplifies to \(|x| < 2\), or \(\mathbf{-2 < x < 2}\).
7. Series Expansion of Rational Functions (P2.2)
Sometimes, you need to find the series expansion of a more complicated algebraic fraction, such as those derived from Partial Fractions.
If you have a function like \(f(x) = \frac{3+2x^2}{(2x+1)(x-3)^2}\), you would first use the Partial Fractions technique (covered in the Algebra chapter) to break it down into simpler components.
Example decomposition: $$\frac{3+2x^2}{(2x+1)(x-3)^2} = \frac{A}{2x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$
Once decomposed, each simple fraction must be rewritten into the form \((1+X)^n\) using negative powers and factorizing constants, as shown in Section 6. Then, apply the Binomial Series expansion to each term separately.
If the power \(n\) is not a positive integer, you must use the infinite series formula and always determine the validity range \(|X| < 1\). If you are combining expansions, the final answer is only valid in the most restrictive range (the intersection of all individual validity ranges).