# Sum and Series and Method of Difference

#### Information

We can represent even, odd and square numbers like below in sum and series equations:
Even Numbers = 2n
Odd Numbers = 2n+1 or 2n-1
Square Numbers = n²

We use the sum symbol $$\sum$$ in order to show that we are adding all numbers within the sum
The example below shows the sum of the numbers 1 to 5
This is produced by first stating the initial value of r which in this case is 1 so 1 is our first number
Then we increase the value of r by 1 for the next iteration up to 2 and add this number on
We do this until we reach the iteration limit which is stated above the sum symbol e.g.5
We then need to add of these iterations together to find the value of the sum which is our answer

$$\sum_{r=1}^{5}r = 1+2+3+4+5 = 15$$

We can also change what we are adding for example 2r means all values of r are doubled
This does not change the amount of iterations as we still add up till r=5

$$\sum_{r=1}^{5}2r = 2+4+6+8+10 = 30$$

We can multiply the value of the sum outside the sum itself like below as well
This means we first find the sum and then multiply it like:

$$2\sum_{r=1}^{5}r = 2(1+2+3+4+5) = 2(15) = 30$$

It is also possible to sum more than just r such as r+1 but the amount of iterations stays the same
Always remeber to put the value you are adding e.g.(2r+1) in brackets if there are multiple parts otherwise it could be misread as you wanting to add all values of 2r and then add 1 to that value afterwards

$$\sum_{r=1}^{5}(2r+1) = 3+5+7+9+11 = 35$$

The inital value of r can also change for instance if r=0 then we start from a value of r=0
Notice there are now 6 iterations as r starts on 0 and needs to reach 5

$$\sum_{r=0}^{5}(2r+1) = 1+3+5+7+9+11 = 36$$

We can increase the limit of r so for instance the new limit is 10 therfore we will have more iterations

$$\sum_{r=0}^{10}(2r+1) = 1+3+5+7+9+11+13+15+17+19+21 = 121$$

The values we are summing can also be squared such as:

$$\sum_{r=1}^{5}(r^2) = 1+4+9+16+25 = 55$$

We can also square the value of the summation which is not the same as squaring everything we add

$$(\sum_{r=1}^{5}r)^2 = (1+2+3+4+5)^2 = 15^2 = 225$$

Below is an important expression you should memorise as it allows you to easily sum all the values with a high iteration number

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

Using the same sum as before in this example we can check that this expression works

$$\sum_{r=1}^{5}r = \frac{1}{2} ×5×(5+1) = \frac{1}{2}×30 = 15$$

This proves that this expression works which means we can now use it for bigger values like if we had 1000 iterations

$$\sum_{r=1}^{1000}r = \frac{1}{2} ×1000×(1000+1) = \frac{1}{2}×1001000 = 500500$$

What if we have an expression from before though where we are not simply summing r but (2r+1)?
In this case we need to apply some logic to the question
We know that when we summed 2r the value of our answer was doubled this means we can easily solve thos problem by doubling our answer for r

But what about the +1 part of the equation
For this we need to realise we are adding 1 to the sum for the amount of iterations there are as when there are 5 iterations the value of 5 is added
This means we can simply add n to the equation or c×n depending on the constant(c)

An example is shown below:

$$\sum_{r=1}^{50}(2r+1) = 2×\frac{1}{2}n(n+1) + n = 2×\frac{1}{2}×50×(50+1) + 50 = 50×(50+1) + 50 = 2550 + 50 = 2600$$

What about if we started with iteration 0?
We can seperate the sum out so we can still sum all iterations from 1 to 50 and then add on the 0th iteration ourselves afterwards like below:

$$\sum_{r=0}^{50}(2r+1) = \sum_{r=1}^{50}(2r+1) + 1 = 2×\frac{1}{2}n(n+1) + n + 1 = 2×\frac{1}{2}×50×(50+1) + 50 + 1 = 50×(50+1) + 51 = 2550 + 51 = 2601$$

(c is a constant e.g.1)

$$\sum_{r=1}^{n}c = c×n$$

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

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

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

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

#### Question 1

What is (1×2)+(2×3)+(3×4)+...+(30×31)?

First we need to find a sum that can represent this equation
As we can see each bracket is a number multiplied by the next
Therefore if we call the first term in the bracket r then r is multiplied by r+1
We can see that the start of each bracket is 1 larger than the last so r is incremented by 1 each time
The first value of r that we have is 1 and the last value is 30, so using all this knowledge we can re-write the equation as:

$$\sum_{r=1}^{30}[r(r+1)] = \sum_{r=1}^{30}[r^2+r]$$

From the helpful equations section we know this can be re-written as:

$$\sum_{r=1}^{30}r^2 + \sum_{r=1}^{30}r = \frac{1}{6}n(n+1)(2n+1) + \frac{1}{2}n(n+1)$$

Inputting our values for n we get:

$$\frac{1}{6}×30×(30+1)(60+1) + \frac{1}{2}×30×(30+1) = \frac{1}{6}×30×(31)(61) + \frac{1}{2}×30×(31) = 9455 + 465 = 9920$$

#### Question 2

What is (6×7)+(7×8)+(8×9)+...+(30×31)?

First we need to find a sum that can represent this equation
As we can see each bracket is a number multiplied by the next
Therefore if we call the first term in the bracket r then r is multiplied by r+1
We can see that the start of each bracket is 1 larger than the last so r is incremented by 1 each time
The first value of r that we have is 6 and the last value is 30, so using all this knowledge we can re-write the equation as:

$$\sum_{r=6}^{30}[r(r+1)] = \sum_{r=6}^{30}[r^2+r]$$

We need to find a way so that we can get this in a term that we can use an equation for
to do this we must find the values of iterations 1 to 30 and then subtract the iterations 1 to 5 as they equal the same value lik so:

$$\sum_{r=6}^{30}(r^2+r) = \sum_{r=1}^{30}(r^2+r) - \sum_{r=1}^{5}(r^2+r)$$

From the previous question we know the first 30 iterations add up to 9920 so we don't need to recalculate this, therefore:

$$9920 - \sum_{r=1}^{5}(r^2+r) = 9920 - [\frac{1}{6}×5×(5+1)(10+1) + \frac{1}{2}×5×(5+1)]$$

$$9920 - \sum_{r=1}^{5}(r^2+r) = 9920 - [\frac{1}{6}×5×(6)(11) + \frac{1}{2}×5×(6)] = 9920 - [55 + 15] = 9850$$

#### Question 3

Simplify:

$$\sum_{r=1}^{n}(6r^2+2r+4) = \frac{6}{6}n(n+1)(2n+1) + \frac{2}{2}n(n+1) + 4n = n(n+1)(2n+1) + n(n+1) + 4n$$

$$n[(n+1)(2n+1) + (n+1) + 4] = n[2n^2+2n+n+1 + n+1 + 4] = n[2n^2+4n+6] = 2n[n^2+2n+3]$$

#### Question 4

Simplify:

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

$$\frac{n}{2}[2(n^3+2n^2+n) - 3(n+1) - 2(4)] = \frac{n}{2}[2n^3+4n^2+2n - 3n-3 - 8] = \frac{n}{2}[2n^3+4n^2-n-11]$$

#### Question 5

Simplify:

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

$$= \frac{n}{6}[(n+1)(2n+1) +9(n+1) + 12] = \frac{n}{6}[2n^2+3n+1 +9n+9 + 12]$$

$$= \frac{n}{6}[2n^2+12n+22] = \frac{n}{3}[n^2+6n+11]$$

#### Question 6

Simplify:

$$\sum_{r=1}^{n}[r(r+1)(r+2)] = \sum_{r=1}^{n}(r^3+3r^2+2r) = \frac{1}{4}n^2(n+1)^2 + \frac{3}{6}n(n+1)(2n+1) + \frac{2}{2}n(n+1)$$

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

$$= \frac{n}{4}[n^3+2n^2+n + 4n^2+6n+2 + 4n+4] = \frac{n}{4}[n^3+6n^2+11n + 6] = \frac{n}{4}(n+1)(n+2)(n+3)$$

#### Method of Difference

In order for a summation to be classed as a Method of Difference there must be a negative sign in the sum

$$e.g.\sum_{r=1}^{n}(\frac{1}{r}-\frac{1}{r+1})$$

Method of Difference is used to give a simpler form of the summation where the answer for any value of n can be easily found
An example of this is shown below:

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

As we can see from this equation everything but the first and last term cancel out this leaves us with the equation:

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

Now for any value of n we can quickly find the answer without the need to do a long calculation every time e.g.n = 30

$$\sum_{r=1}^{30}(\frac{1}{r}-\frac{1}{r+1}) = \frac{30}{30+1} = \frac{30}{31}$$

Another examplee of Method of Difference is as follows where the pattern is harder to spot:

$$\sum_{r=1}^{30}(\frac{2}{r}-\frac{3}{r+1}+\frac{1}{r+2})$$

$$\frac{2}{1}-\frac{3}{2}+\frac{1}{3}+\frac{2}{2}-\frac{3}{3}+\frac{1}{4}+\frac{2}{3}-\frac{3}{4}+\frac{1}{5}+\frac{2}{4}-\frac{3}{5}+\frac{1}{6}+...$$

$$-...+\frac{2}{n-2}-\frac{3}{n-1}+\frac{1}{n}+\frac{2}{n-1}-\frac{3}{n}+\frac{1}{n+1}+\frac{2}{n}-\frac{3}{n+1}+\frac{1}{n+2}$$

Now if we collect all the fractions with the same denominators together we see that anything with a denominator of 3 or greater cancels out
Now we just need to think about the last few terms in the sequence will be
If we look at the set of terms we created we see after cancelling we are left with
(We can assume all previous n denominator fractions cancel as this trend has been set at the start of the sequence e.g.the denominator of 3 cancels out earlier)

$$\frac{2}{1}-\frac{3}{2}+\frac{2}{2}+\frac{1}{n+1}-\frac{3}{n+1}+\frac{1}{n+2} = \frac{2}{1}-\frac{1}{2}-\frac{2}{n+1}+\frac{1}{n+2}$$

Therefore our final equation becomes:

$$\frac{3}{2}-\frac{2}{n+1}+\frac{1}{n+2}$$

This final example uses partial fractions which we will learn about next so if you do not know anything about partial fractions please read that first and then come back

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

Finding values for the first few and last terms we get:

$$\frac{1}{3}(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}+...-\frac{1}{3n+2})$$

We can see that all of the middle terms cancel out so we are only left with the first and last terms

Remembering to multiply throughout by 1/3 we are left with the equation:

$$\frac{1}{6}-\frac{1}{9n+6}$$