# Partial Sums

A Partial Sum is a **Sum** of **Part** of a Sequence.

### Example:

This is the **Sequence** of even numbers: {2, 4, 6, 8, 10, 12, ...}

This is the **Partial Sum** of the first 4 terms of that sequence: 2+4+6+8 = 20

Let us define things a little better now:

A **Sequence** is a set of things (usually numbers) that are in order.

A **Partial Sum** is the sum of **part** of the sequence

Language Note: Partial Sums are sometimes called **"Finite Series"**

(a **"Series"** is the sum of an infinite sequence).

*(Note: The sum of infinite terms is an Infinite Series.)*

## Sigma

Partial Sums are often written using Σ to mean "add them all up":

This symbol (called Sigma) means "sum up" |

So | means to sum things up ... |

## Sum What? |
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Sum whatever appears after the Sigma: |
so we sum |
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## But What is the Value of n |
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The values are shown below |
it says |
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## OK, Let's Go ... |
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So now we add up 1,2,3 and 4: |

Here it is in one diagram:

## More Powerful

But Σ can do more powerful things than that!

We could square *n* each time and sum the result:

We could "add up the first four terms in the sequence **2n+1**":

And we don't have to use *n*. Here we use i and sum up i × (i+1), going from **1** to **3**:

And we can start and end with any number. Here we go from **3** to **5**:

## Properties

Partial Sums have some useful properties that can help us do the calculations.

### Multiplying by a Constant Property

Say we have something we want to sum up, let's call it a_{k}

a_{k} could be **k ^{2}**, or

**k(k-7)+2**, or ... anything really

And **c** is some constant value (like **2**, or **-9.1**, etc), then:

In other words: if every term we are summing is multiplied by a constant, we can "pull" the constant outside the sigma.

### Example:

So instead of summing **6k ^{2}** we can sum

**k**and then multiply the whole result by

^{2}**6**

### Adding or Subtracting Property

Here is another useful fact:

Which means that when two terms are added together, and we want to sum them up, we can actually sum them **separately** and then **add **the results.

### Example:

It is going to be easier to do the two sums and then add them at the end.

Note this also works for subtraction:

## Useful Shortcuts

And here are some useful shortcuts that make the sums **a lot easier**.

In each case we are trying to sum from **1** to some value **n**.

Summing 1 equals n |
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Summing the constant c equals c times n |
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A shortcut when summing k |
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A shortcut when summing k^{2} |
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A shortcut when summing k^{3} |

Let's use some of those:

### Example 1: You sell concrete blocks for landscaping.

A customer says they will buy the entire "pyramid" of blocks you keep out front. The stack is 14 blocks high.

How many blocks are in there?

Each layer is a square, so the calculation is:

1^{2} + 2^{2} + 3^{2} + ... + 14^{2}

But this can be written **much more easily** as:

We can use the formula for **k ^{2}** from above:

That was a lot easier than adding up **1 ^{2} + 2^{2} + 3^{2} + ... + 14^{2}**.

And here is a more complicated example:

### Example 2: The customer wants a better price.

The customer says the blocks on the outside of the pyramid should be cheaper, as they need cleaning.

You agree to:

- $7 for outer blocks
- and $11 for inner blocks.

What is the total cost?

You can calculate how many "inner" and "outer" blocks in any layer (except the first) using

- outer blocks =
**4×(size-1)** - inner blocks =
**(size-2)**^{2}

And so the cost per layer is:

- cost (outer blocks) = $7 × 4(size-1)
- cost (inner blocks) = $11 × (size-2)
^{2}

So all layers together (except first) will cost:

Now we have the sum, let us try to make the calculations easier!

Using the "Addition Property" from above:

Using the "Multiply by Constant Property" from above:

That is good ... but we can't use any shortcuts as it is, as we are going from **i=2** instead of** i=1**

HOWEVER, if we invent two new variables:

- j = i-1
- k = i-2

We have:

(I dropped the k=0 case, because I know that 0^{2}=0)

And now we can use the shortcuts:

After a little calculation:

$7 × 364 + $11 × 650 = $9,698.00

Oh! And don't forget the top layer (size=1) which is just one block. Maybe you can give them that one for free, you are so generous!

*Note: as a check, when we add the "outer" and "inner" blocks, plus the one on top, we get*

** 364 + 650 + 1 = 1015**

*Which is the same number we got for the "total blocks" before ... yay!*