Series

Q: What are the two linearity properties used to manipulate sums?

1. Constant Rule: $\sum k f(r) = k \sum f(r)$. 2. Addition/Subtraction Rule: $\sum [f(r) \pm g(r)] = \sum f(r) \pm \sum g(r)$.

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Further Mathematics (YFM01) · Pearson Edexcel International A Level · 7 min read · Updated: 7 Sept 2026

Welcome to FP1 Series: Summing Up Success!

Hello future Further Mathematicians! This chapter, Series, is all about finding fast and efficient ways to calculate the sum of many terms without having to add them up one by one. Think of it like a mathematical shortcut!

Why is this important? Because series pop up everywhere—from calculating compound interest to analyzing how quickly a computer algorithm runs. By mastering these techniques, you'll gain powerful tools for dealing with complex sums quickly and accurately. Don't worry if this seems tricky at first; we'll break it down into manageable steps.

Quick Review: The Sigma Notation ($\Sigma$)

Before we jump into the big formulas, let's quickly review the essential language we use: Sigma Notation.

The Greek letter $\Sigma$ (Sigma) simply means "sum."

The expression $\sum_{r=1}^{n} f(r)$ means: "Add up all the terms $f(r)$ starting from $r=1$ (the lower limit) up to $r=n$ (the upper limit)."

Key Properties of Sigma Notation (Linearity)

These properties are vital when manipulating complex sums:

Constant Rule: A constant factor can be pulled outside the sum.
$\sum_{r=1}^{n} k \cdot f(r) = k \sum_{r=1}^{n} f(r)$
Addition/Subtraction Rule: The sum of separate terms can be split into separate sums.
$\sum_{r=1}^{n} [f(r) + g(r)] = \sum_{r=1}^{n} f(r) + \sum_{r=1}^{n} g(r)$

Key Takeaway: Sigma notation is just a compact way of writing a long sum. Use the linearity properties to separate complicated expressions into easier sums.

Section 1: Sums of Powers of Integers

The core of the FP1 series chapter involves memorizing and correctly applying three standard formulae for the sum of the first $n$ positive integers, their squares, and their cubes. These results are often provided in the formula booklet, but familiarity speeds up your exams significantly!

The Three Standard Summation Formulae

1. Sum of the First $n$ Integers (Arithmetic Series)

This is the sum of $1 + 2 + 3 + \dots + n$. We denote this as $S_1$.

$$ \sum_{r=1}^{n} r = \frac{1}{2} n (n+1) $$

(Did you know? This formula is attributed to the brilliant mathematician Gauss, who supposedly discovered it as a child!)

2. Sum of the First $n$ Squares

This is the sum of $1^2 + 2^2 + 3^2 + \dots + n^2$. We denote this as $S_2$.

$$ \sum_{r=1}^{n} r^2 = \frac{1}{6} n (n+1) (2n+1) $$

Memory Aid Tip: Notice the fraction $\frac{1}{6}$ and the term $(2n+1)$. If you forget the $\frac{1}{6}$, remember that $1 \times 2$ is the base of the fraction, and you need the $2n+1$ factor to account for the square nature of the sum.

3. Sum of the First $n$ Cubes

This is the sum of $1^3 + 2^3 + 3^3 + \dots + n^3$. We denote this as $S_3$.

$$ \sum_{r=1}^{n} r^3 = \frac{1}{4} n^2 (n+1)^2 $$

THE BEST MEMORY TRICK: Look closely at the formula for $S_3$. It is simply the square of the formula for $S_1$!

$$ \sum_{r=1}^{n} r^3 = \left[ \sum_{r=1}^{n} r \right]^2 $$

This connection makes $S_3$ the easiest formula to remember!

Quick Review: Standard Sums

$\sum r = \frac{1}{2} n (n+1)$
$\sum r^2 = \frac{1}{6} n (n+1) (2n+1)$
$\sum r^3 = \left[ \sum r \right]^2$

Section 2: Solving Polynomial Series Problems

In exams, you won't just be asked to sum $r^2$. You will be asked to sum a polynomial in $r$, such as $\sum_{r=1}^{n} (3r^2 - 4r + 2)$.

The process requires two steps: 1) Separation and 2) Substitution/Simplification.

Step-by-Step Process for Polynomial Sums

Step 1: Expand and Separate the Sum

Use the linearity properties (from the review section) to break the complex sum into individual standard sums.

Example: Find $\sum_{r=1}^{n} (3r^2 - 4r + 2)$

$$ \sum_{r=1}^{n} (3r^2 - 4r + 2) = 3 \sum_{r=1}^{n} r^2 - 4 \sum_{r=1}^{n} r + \sum_{r=1}^{n} 2 $$

Step 2: Handle the Constant Term

Be very careful with constant terms (like $\sum 2$). This means summing the number 2, $n$ times.

$$ \sum_{r=1}^{n} c = c + c + c + \dots + c \quad (n \text{ times}) = cn $$

In our example, $\sum_{r=1}^{n} 2 = 2n$.

Step 3: Substitute and Factorise

Substitute the standard formulae into your separated expression.

$$ 3 \left( \frac{1}{6} n(n+1)(2n+1) \right) - 4 \left( \frac{1}{2} n(n+1) \right) + 2n $$

$$ = \frac{1}{2} n(n+1)(2n+1) - 2 n(n+1) + 2n $$

The final step is crucial: factor out common terms, usually $n$ and common fractions, to simplify the expression into a neat polynomial.

In the example above, factoring out $n$ gives: $$ = n \left[ \frac{1}{2} (n+1)(2n+1) - 2 (n+1) + 2 \right] $$

Continue simplifying the bracket until you have a fully factorized answer. This step often involves dealing with fractions (e.g., finding a common denominator like $\frac{1}{2}$ or $\frac{1}{3}$).

Common Mistake Alert: Sums Not Starting at $r=1$

If you are asked to find the sum from $r=k$ to $n$ (where $k > 1$), you CANNOT use the standard formulas directly.

$$ \sum_{r=k}^{n} f(r) = \sum_{r=1}^{n} f(r) - \sum_{r=1}^{k-1} f(r) $$

Example: Find $\sum_{r=5}^{10} r^2$.

You calculate the sum for the first 10 terms ($n=10$) and subtract the sum for the first 4 terms ($n=4$). We subtract up to $k-1 = 5-1 = 4$.

$$ \sum_{r=5}^{10} r^2 = \sum_{r=1}^{10} r^2 - \sum_{r=1}^{4} r^2 $$

Key Takeaway: To solve polynomial series, separate the terms, use the standard formulas, and factorise the final result completely. Remember to adjust when the sum doesn't start at $r=1$.

Section 3: The Method of Differences (Telescoping Series)

Polynomial series are solved using fixed formulas. However, some series, particularly those involving fractions (rational functions), cannot be solved this way. For these, we use the Method of Differences, often called a Telescoping Series.

What is a Telescoping Series?

Imagine an old pirate telescope: when you collapse it, the sections slide into each other, leaving only the two end pieces visible. That's exactly what happens mathematically! Most of the middle terms cancel each other out.

The Fundamental Principle

For the method to work, the term $f(r)$ must be expressed as a difference between two consecutive terms of a sequence, $g(r)$.

$$ f(r) = g(r) - g(r+1) \quad \text{OR} \quad f(r) = g(r+1) - g(r) $$

Step-by-Step Process for Method of Differences

Step 1: Express $f(r)$ as a Difference

If $f(r)$ is a rational function (a fraction), you will usually need to use Partial Fractions to decompose it into the required difference form.

Example: Find the sum of $\sum_{r=1}^{n} \frac{1}{r(r+1)}$

Using partial fractions: $$ \frac{1}{r(r+1)} = \frac{A}{r} + \frac{B}{r+1} $$

You find $A=1$ and $B=-1$.

So, $f(r)$ can be written as: $$ f(r) = \frac{1}{r} - \frac{1}{r+1} $$

Here, $g(r) = \frac{1}{r}$ and $g(r+1) = \frac{1}{r+1}$.

Step 2: Write Out the Sum and Observe the Cancellation

Write out the first few terms and the last few terms of the series:

$r=1$: $\left( \frac{1}{1} - \frac{1}{2} \right)$
$r=2$: $\left( \frac{1}{2} - \frac{1}{3} \right)$
$r=3$: $\left( \frac{1}{3} - \frac{1}{4} \right)$
...
$r=n-1$: $\left( \frac{1}{n-1} - \frac{1}{n} \right)$
$r=n$: $\left( \frac{1}{n} - \frac{1}{n+1} \right)$

When you add these terms together, notice the cancellation: $$ S_n = \left( \frac{1}{1} \cancel{- \frac{1}{2}} \right) + \left( \cancel{\frac{1}{2}} \cancel{- \frac{1}{3}} \right) + \left( \cancel{\frac{1}{3}} \cancel{- \frac{1}{4}} \right) + \dots + \left( \cancel{\frac{1}{n-1}} \cancel{- \frac{1}{n}} \right) + \left( \cancel{\frac{1}{n}} - \frac{1}{n+1} \right) $$

Step 3: State the Remaining Terms

Only the first part of the $r=1$ term and the last part of the $r=n$ term remain!

$$ S_n = \frac{1}{1} - \frac{1}{n+1} = 1 - \frac{1}{n+1} $$

You can then combine this final expression over a common denominator if required: $$ S_n = \frac{n+1 - 1}{n+1} = \frac{n}{n+1} $$

Important Variations and Caution

Sometimes, the difference isn't between $g(r)$ and $g(r+1)$, but between terms further apart, like $g(r)$ and $g(r+2)$.

If $f(r) = g(r) - g(r+2)$:

You write out four terms ($r=1, 2, n-1, n$).
The cancellation skips one term.
The remaining terms will be the first two positive terms and the last two negative terms.
$$ S_n = [g(1) + g(2)] - [g(n+1) + g(n+2)] $$

Encouragement: Don't panic if the cancellation looks complex. Always write out the first two and last two lines explicitly. Draw lines through the canceling terms—this visual method is the clearest way to avoid errors!

Key Takeaway: The Method of Differences relies on expressing the general term $f(r)$ as a difference $g(r) - g(r+k)$. When summed, only a few terms at the beginning and end survive the cancellation.

Final Summary and Encouragement

You have now covered the two main pillars of the FP1 Series chapter:

Standard Sums: Used for polynomial expressions in $r$. Requires separation, substitution of $\sum r$, $\sum r^2$, or $\sum r^3$, and rigorous factorisation.
Method of Differences: Used typically for rational expressions (fractions). Requires partial fractions to create a difference $g(r) - g(r+k)$ and relies on writing out the series to show cancellation (telescoping).

Practice is essential here, especially the factorization steps for the standard sums, and the partial fractions/cancellation steps for the method of differences. You can do this! Good luck!

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What does the notation $\sum_{r=1}^{n} f(r)$ represent in the context of series?

It is Sigma notation representing the sum of all terms generated by the function $f(r)$, starting with $r=1$ and ending at $r=n$.

What are the two linearity properties used to manipulate sums?

1. Constant Rule: $\sum k f(r) = k \sum f(r)$. 2. Addition/Subtraction Rule: $\sum [f(r) \pm g(r)] = \sum f(r) \pm \sum g(r)$.

What is the standard formula for the sum of the first $n$ positive integers?

$\sum_{r=1}^{n} r = \frac{1}{2} n (n+1)$

What is the standard formula for the sum of the first $n$ squares?

$\sum_{r=1}^{n} r^2 = \frac{1}{6} n (n+1) (2n+1)$

What is the standard formula for the sum of the first $n$ cubes?

$\sum_{r=1}^{n} r^3 = \frac{1}{4} n^2 (n+1)^2$

How is the sum of the first $n$ cubes related to the sum of the first $n$ integers?

The sum of the first $n$ cubes is exactly equal to the square of the sum of the first $n$ integers: $\sum r^3 = [\sum r]^2$.

What is the result when summing a constant value $c$ from $r=1$ to $n$?

$\sum_{r=1}^{n} c = cn$

What is the final critical step when evaluating polynomial sums algebraically?

The final step is to factorise the expression fully, typically by pulling out common factors like $n$ and common denominators.

How do you calculate a sum where the lower limit is not 1 (e.g., from $r=k$ to $n$)?

$\sum_{r=k}^{n} f(r) = \sum_{r=1}^{n} f(r) - \sum_{r=1}^{k-1} f(r)$

What is the fundamental principle of the Method of Differences?

It involves expressing the general term $f(r)$ as a difference between two terms of a sequence, $g(r) - g(r+1)$, so that intermediate terms cancel out.

How are partial fractions used in the Method of Differences?

Partial fractions are used to split a single rational term (a fraction) into two or more simpler fractions that form the required difference $g(r) - g(r+k)$.

What happens in the Method of Differences if the terms are 'far apart', such as $g(r) - g(r+2)$?

The cancellation skips a term, resulting in two terms remaining at the start and two terms remaining at the end of the sum.

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