9231. Further Pure 1. Summation of Series

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, Σr2, Σr3

Let’s think about language first.

$\sum\limits_{r=1}^{n} k$ means that k is added together n times, i.e. this is equal to kn.

However, $\sum\limits_{r=1}^{n} r$ 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 $\frac{n(n+1)}{2}$

In summary, $\sum\limits_{r=1}^{n} r =\frac{n(n+1)}{2}$

Worked Example

Find $\sum\limits_{r=1}^{n} (4r+1)$ and hence calculate $\sum\limits_{17}^{48} (4r+1)$

Note

We are not restricted to “n” as our upper limit, for instance $\sum\limits_{r=1}^{2n} r$ would be $\frac{2n(2n+1)}{2}=2n^2+n$

We can use this to deduce $\sum\limits_{n+1}^{2n} r$ as this is the sum of the first 2n terms minus the first n terms, so is $\sum\limits_{1}^{2n} r - \sum\limits_{1}^{n}r$, which is $2n^2 + n - \frac{n(n+1)}{2} = \frac{3}{2}n^2 + \frac{1}{2}n$

Worked example 2

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

Σr2

Let’s work with the binomial expansion (r+1)3=r3+3r2+3r+1

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

23 = 13 + 3×12 + 3×1 + 1

33 = 23 + 3×22 + 3×2 + 1

43 = 33 + 3×32 + 3×3 +1

(n+1)3 = n3 + 3xn2 +3xn +1

If we add all of the terms on the right hand side, we have $\sum\limits_{r=1}^{n} r^3 + 3 \sum\limits_{r=1}^{n}r^2 + 3 \sum\limits_{r=1}^{n}r +n$

We can then add 1 (i.e. 13) to both aides of the equation to give us:

$\sum\limits_{r=1}^{n+1} r^3=\sum\limits_{r=1}^{n} r^3 + 3 \sum\limits_{r=1}^{n}r^2 + 3 \sum\limits_{r=1}^{n}r +n+1$

We notice that both sides of the equation include $\sum\limits_{r=1}^{n} r^3$ so we can subtract this leaving $(n+1)^3= 3 \sum\limits_{r=1}^{n}r^2 + 3 \sum\limits_{r=1}^{n}r +n+1$ which we can then solve to give a formula for $\sum\limits_{r=1}^{n}r^2$

Worked Example

Find an expression in terms of n for $\sum\limits_{r=1}^{n}(3r^2-4r+2)$

Σr3

Similar to our last example, let’s start with the binomial expansion (r+1)4=r4+4r3+6r2+4r+1, and substitute in successive natural numbers, giving:

24 = 14 + 4×13 + 6×12 + 4×1 + 1

34 = 24 + 4×23 + 6×22 + 4×2 + 1

44 = 34 + 4×33 + 6×32 + 4×3 +1

(n+1)4 = n4 + 4xn3 + 6xn2 +4xn +1.

Adding 1 to both sides and summing give:

$\sum\limits_{r=1}^{n+1} r^4=\sum\limits_{r=1}^{n} r^4 + 4 \sum\limits_{r=1}^{n}r^3 + 6 \sum\limits_{r=1}^{n}r^2 + 4 \sum\limits_{r=1}^{n}r +n+1$

Following the same procedures as above, we can rearrange this to find $\sum\limits_{r=1}^{n} r^3$

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

Worked Example

Find an expression in terms of n for $\sum\limits_{r=1}^{n} r^2(r-1)$

Exercise 1

Answers to Exercise 1

1,2. Proof; 3. -n, 4. Proof; 5. $\frac{1}{2}n(6n^2-9n-5)$ 6. $\frac{1}{3}n(2+9n+7n^2)$, 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 $\sum\limits_{r=1}^{n} \frac{1}{r(r+1)}$.

If we look at the first few terms, $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20}$ it is hard to spot a pattern.

So let’s try a different methods. Using the partial fractions method, let’s rewrite $\frac{1}{r(r+1)}$ (do you remember how to do this?)

So we can now look at $\sum\limits_{r=1}^{n} (\frac{1}{r}-\frac{1}{r(r+1)})$.

If we look at the first few terms of this: $(\frac{1}{1}-\frac{1}{2}) + (\frac{1}{2}-\frac{1}{3}) + (\frac{1}{3}-\frac{1}{4}) + ... + (\frac{1}{n-1}-\frac{1}{n}) + (\frac{1}{n} - \frac{1}{n+1})$ we see that all of the terms in the middle cancel out, leaving us with just $1 - \frac{1}{n+1}$, i.e. $\frac{n}{n+1}$

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
• $\sum\limits_{r=1}^{n} (\frac{1}{(r+1)(r+2)})$;
• $\sum\limits_{r=1}^{n} (\frac{1}{r(r+2)})$;
• $\sum\limits_{r=1}^{n} (\frac{1}{(r-1)(r+1)})$;
• $\sum\limits_{r=2}^{n} (\frac{5}{r(r-1)})$.

Exercise 2 and General Exercise

Answers to Exercise 2 and General Exercise

Worked Solutions to Exercise 2

Past Paper Exam Questions

Summer 2020 11:

Summer 2020 11 mark scheme

Winter 2020 11:

Winter 2020 11 mark scheme

Winter 2020 12:

Winter 2020 12 mark scheme

Summer 2019 11:

Summer 2019 11 Mark Scheme

Summer 2019 13:

Summer 2019 13 Mark Scheme