Define the method of differences .

It is a technique for summing a series whose general term can be written as a difference , so that almost every term is cancelled by part of another term. Only a few terms at the very start and the very end survive, which turns a long summation into a short expression.

The general term of a series is a single algebraic fraction. What has to be done before the method of differences can be used on it?

Split it into partial fractions , so that the term becomes a difference of two or more simpler fractions. Until that is done there is nothing for the cancelling to act on, because a single fraction has no parts that can cancel against its neighbours.

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Series (Edexcel A Level Further Maths: Core Pure): Flashcards

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FrontSums of Integers, Squares & Cubes
Define the sigma notation $\sum_{r = 1}^{n} u_{r}$ .

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Define the sigma notation $\sum_{r = 1}^{n} u_{r}$ .
It stands for the sum of every term of the sequence $u_{r}$ , from $r = 1$ up to $r = n$ inclusive.
The letter beneath the sigma is the variable of summation, and the values above and below it are the limits between which that variable runs.
Define the method of differences.
It is a technique for summing a series whose general term can be written as a difference, so that almost every term is cancelled by part of another term.
Only a few terms at the very start and the very end survive, which turns a long summation into a short expression.
Complete the formula for the sum of the first $n$ natural numbers:
$\sum_{r = 1}^{n} r =______$
The completed formula is:
$\sum_{r = 1}^{n} r = \frac{1}{2} n (n + 1)$
This is the sum of the arithmetic series $1 + 2 + \dots + n$ , which is $\frac{1}{2} n$ times the sum of the first and last terms.
The general term of a series is a single algebraic fraction. What has to be done before the method of differences can be used on it?
Split it into partial fractions, so that the term becomes a difference of two or more simpler fractions.
Until that is done there is nothing for the cancelling to act on, because a single fraction has no parts that can cancel against its neighbours.
What is $\sum_{r = 1}^{n} a$ , where $a$ is a constant?
It is $n a$ , since the constant is added to itself once for each of the $n$ values of $r$ .
The variable $r$ never appears in the term, so nothing changes from one term to the next and the sum is just a repeated addition.
Complete the result of the simplest method of differences sum:
$\sum_{r = 1}^{n} (f (r) - f (r + 1)) =______-______$
The completed result is:
$\sum_{r = 1}^{n} (f (r) - f (r + 1)) = f (1) - f (n + 1)$
Every other term appears twice, once positive and once negative, so only the very first and the very last are left standing.
True or False?
$\sum (a f (r) + b g (r))$ can be split into $a \sum f (r) + b \sum g (r)$ whatever limits the summations carry.
False.
The splitting is valid only when every summation involved runs between the same upper and lower limits.
With different limits the two sides are adding up different numbers of terms, so there is no reason for them to be equal.
For a sum of the form $\sum_{r = 1}^{n} (f (r) - f (r + 2))$ , how many terms survive the cancelling, and which are they?
Four survive, namely $f (1) + f (2) - f (n + 1) - f (n + 2)$ .
Each term now cancels against one two rows further down rather than the next one, so two terms are stranded at each end instead of one.
In general the gap in the argument fixes how many are left at each end.
Why is $\sum r^{2}$ not the same as ${(\sum r)}^{2}$ ?
Squaring the whole sum produces all the cross terms as well as the squares, whereas $\sum r^{2}$ adds the squares only.
For $n = 3$ this is the difference between $1 + 4 + 9 = 14$ and ${(1 + 2 + 3)}^{2} = 36$ .
True or False?
The method of differences can be used on a series whose general term is not a fraction at all.
True.
All that is needed is a term that can be written as a difference, and the standard formulae for $\sum r$ , $\sum r^{2}$ and $\sum r^{3}$ are themselves proved this way from differences of powers.
Fractions are simply the commonest case, because partial fractions produce differences so readily.
How is $\sum_{r = 1}^{n} r^{3}$ related to $\sum_{r = 1}^{n} r$ ?
The sum of the cubes is the square of the sum of the natural numbers:
$\sum_{r = 1}^{n} r^{3} = {(\sum_{r = 1}^{n} r)}^{2} = \frac{1}{4} n^{2} {(n + 1)}^{2}$
For example $1^{3} + 2^{3} + 3^{3} = 36$ , which is ${(1 + 2 + 3)}^{2}$ .
When using the method of differences, why do you write out the first and last few terms in full?
Writing them out is what reveals which terms cancel and which are left over, since that pattern cannot be read off the sigma notation on its own.
The closing terms have to be written in terms of $n$ , $n - 1$ and so on, because the terms that survive at the top end are expressed in $n$ .
How do you evaluate a summation whose general term is a product of brackets, such as $\sum (2 r + 1) (r - 3) (r + 1)$ ?
Expand the brackets first, so that the general term becomes a polynomial in $r$ , then split the summation term by term and bring each constant outside its own sigma.
Every piece left is then either one of the standard sums of $r$ , $r^{2}$ or $r^{3}$ , or the sum of a constant.
A sum runs from $r = 20$ instead of from $r = 1$ . How do you use a result that was proved for sums starting at $r = 1$ ?
Subtract off the opening terms you do not want:
$\sum_{r = p}^{q} u_{r} = \sum_{r = 1}^{q} u_{r} - \sum_{r = 1}^{p - 1} u_{r}$
The subtracted sum stops at $p - 1$ and not at $p$ , or the first term you actually wanted would be removed along with the rest.
How can the method of differences prove the formula for $\sum_{r = 1}^{n} r^{2}$ ?
Start from an identity whose difference produces a multiple of $r^{2}$ , namely ${(2 r + 1)}^{3} - {(2 r - 1)}^{3} = 24 r^{2} + 2$ , and sum both sides from $r = 1$ to $n$ .
The left-hand side telescopes down to two terms, and rearranging the equation that is left gives the formula.

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