A-level Mathematics/FP1/Summation of Series

Summation of a Series

In Core Two we learned about arithmetic and geometric progression, but if we need to sum an arithmetic progression over a large range it can become very time consuming. There are formulae that can allow us to calculate the sum. Note that these formulae only work if we start from 1; we will see how to calculate summations from other starting points in the example below. The formulae are:

$\sum _{r=1}^{n}1=n$

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

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

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

We also need to know this general result about summation:

$\sum _{r=1}^{n}a{r}^b=a\sum _{r=1}^n{r}^b$

You can see why this is true by thinking of the expanded form:

$(a\times 1^b+a\times 2^b+a\times 3^b+...+a\times n^b)\equiv a(1^b+2^b+3^b+...+n^b)$

Example

Find the sum of the series $\sum _{x=3}^{10}3x^3+4x^2+5x$.

1. First we need to break the summation into its three separate components.

   $\sum _{x=3}^{10}3x^3+\sum _{x=3}^{10}4x^2+\sum _{x=3}^{10}5x$

2. Next we need to make them start from one. We then need to subtract the sum of the numbers not included in the summation.

   $\sum _{x=1}^{10}3x^3-\sum _{x=1}^{2}3x^3+\sum _{x=1}^{10}4x^2-\sum _{x=1}^{2}4x^2+\sum _{x=1}^{10}5x-\sum _{x=1}^{2}5x$

3. Now we use the identities to calculate the individual sums. Remember to include the co-efficients.

   $3[\frac{1}{4}10^2{(10+1)}^2-\frac{1}{4}{2}^2{(2+1)}^2]+4[\frac{1}{6}10(2\times 10+1)(10+1)-\frac{1}{6}2(2\times 2+1)(2+1)]$ $+5[\frac{1}{2}10(10+1)-\frac{1}{2}2(2+1)]$

4. Now we need to perform a lot of algebra. This can be done by hand or utilizing a calculator.

   $3(\frac{1}{4}\times 100\times 121-\frac{1}{4}4\times 9)+4(\frac{1}{6}\times 10\times 21\times 11-\frac{1}{6}\times 10\times 4)$ $+5(\frac{1}{2}\times 10\times 11-\frac{1}{2}\times 2\times 3)$

   $=(9075-27)+(385-5)+(275-15)$

   $=10828$

5. The sum of the series $\sum _{x=3}^{10}3x^3+4x^2+5x=10828$.

Method of Differences

The method of differences works when the resultants of the series cancel out in pairs leaving only two resultants. The method of differences only works with:

• Polynomial identities
• Polynomials in the denominator, which can be put into partial fractions (which are discussed in Core 4)
• Trigonometric functions

Suppose we wish to find:

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

This can be expressed as:

$\sum _{r=1}^n\frac{1}{r(r+1)}-\frac{1}{(r+1)(r+1)}$

We expand a few terms of the series at the beginning and end:

$[\frac{1}{2}-\frac{1}{6}]+[\frac{1}{6}-\frac{1}{12}]+[\frac{1}{12}-\frac{1}{20}]+\dots +[\frac{1}{(n-1)n}-\frac{1}{n(n+1)}]+[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}]$

As you can see, all the terms except the first and last cancel out, leaving us with:

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

To find:

$\sum _{r=1}^{\mathrm{\infty }}\frac{2}{r(r+1)(r+2)}$

We simply observe that:

$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{(n+1)(n+2)}=0$

Hence the sum to infinity is simply $\frac{1}{2}$. There is a general rule about this method:

In the example, f(r) was:

$-\frac{1}{r(r+1)}$