A **series** is the sum of all the terms in a sequence (the sequence may be **finite** or **infinite**).

You have already met arithmetic and geometric series and applied the formulae for their series:

We will build on and extend this work, by looking at **convergent **series and series of squares and cubes of numbers.

**Summation formulae: Σr, Σr ^{2}, Σr^{3}**

Let’s think about language first.

means that k is added together n times, i.e. this is equal to kn.

However, is quite different and means the sum of the digits from 1 to n, i.e. an arithmetic sequence with first term 1 and difference 1, which from the formula above has a sum of

In summary,

**Worked Example**

Find and hence calculate

**Note**

We are not restricted to “n” as our upper limit, for instance would be

We can use this to deduce as this is the sum of the first 2n terms minus the first n terms, so is , which is

**Worked example 2**

Find an expression for the first n terms of: 1 + 3 + 5 + 7 + …

**Σr ^{2}**

Let’s work with the binomial expansion (r+1)^{3}=r^{3}+3r^{2}+3r+1

If we substitute 1,2,3, …,n into this equation we get:

2^{3} = 1^{3} + 3×1^{2} + 3×1 + 1

3^{3} = 2^{3} + 3×2^{2} + 3×2 + 1

4^{3} = 3^{3} + 3×3^{2} + 3×3 +1

…

(n+1)^{3} = n^{3} + 3xn^{2} +3xn +1

If we add all of the terms on the right hand side, we have

We can then add 1 (i.e. 1^{3}) to both aides of the equation to give us:

We notice that both sides of the equation include so we can subtract this leaving which we can then solve to give a formula for

**Worked Example**

Find an expression in terms of n for

**Σr ^{3}**

Similar to our last example, let’s start with the binomial expansion (r+1)^{4}=r^{4}+4r^{3}+6r^{2}+4r+1, and substitute in successive natural numbers, giving:

2^{4} = 1^{4} + 4×1^{3} + 6×1^{2} + 4×1 + 1

3^{4} = 2^{4} + 4×2^{3} + 6×2^{2} + 4×2 + 1

4^{4} = 3^{4} + 4×3^{3} + 6×3^{2} + 4×3 +1

…

(n+1)^{4} = n^{4} + 4xn^{3} + 6xn^{2} +4xn +1.

Adding 1 to both sides and summing give:

Following the same procedures as above, we can rearrange this to find

**Standard Summations Reminder (in formula book, but also good to memorise)**

**Worked Example**

Find an expression in terms of n for

**Exercise 1**

**Answers to Exercise 1**

1,2. Proof; 3. -n, 4. Proof; 5. 6. , 7. 328,350, 8. Proof, 32,760, 9. 204,020,000

Worked Solutions to Exercise 1

**Converging Series**

We will now look at infinite series with a finite sum. In order to identify the finite sum, we will have to spot patterns within the series.

Let’s consider first .

If we look at the first few terms, it is hard to spot a pattern.

So let’s try a different methods. Using the **partial fractions** method, let’s rewrite (do you remember how to do this?)

So we can now look at .

If we look at the first few terms of this: we see that all of the terms in the middle cancel out, leaving us with just , i.e.

When we cancel terms in this way we are using what is called the **method of differences**.

**Worked examples**

- Find in terms of n, the sum of
- ;
- ;
- ;
- .

**Exercise 2 and General Exercise**

**Answers to Exercise 2 and General Exercise**

Worked Solutions to Exercise 2

**Past Paper Exam Questions**

Summer 2020 11:

Winter 2020 11:

Winter 2020 12:

Summer 2019 11:

Summer 2019 13: