# Sum of *k*

How to use the sum of k formula to solve summation problems: formula, proof, example, and its solution.

## Formula

The sum of *k* as *k* goes from 1 to *n*

is *n*(*n* + 1)/2.

## Proof

Previouly, you've proved this formula.

Mathematical induction

So let's see the proof

not only to learn about the formula,

but also to remember

how to write mathematical induction.

Show that

'if *n* = 1, the given statement is true'.

Assume that

'if *n* = *k*, the given statement is true'.

So assume that

1 + 2 + 3 + ... + *k* = *k*(*k* + 1)/2

is true.

Show that

'if *n* = *k* + 1, the given statement is true'.

For *n* = *k* + 1,

(left side) = 1 + 2 + 3 + ... + *k* + (*k* + 1).

So add (*k* + 1) on both sides:

1 + 2 + 3 + ... + *k* + (*k* + 1) = *k*(*k* + 1)/2 + (*k* + 1).

Then change the right side

to (*k* + 1)[(*k* + 1) + 1]/2.

By showing this, you can prove that

'if *n* = *k* + 1, the given statement is true'.

This part is the most tricky part

when solving mathematical induction problems.

'If *n* = 1, the given statement is true' makes

'if *n* = 2, the given statement is true'.

'If *n* = 2, the given statement is true' makes

'if *n* = 3, the given statement is true'.

'If *n* = 3, the given statement is true' makes

'if *n* = 4, the given statement is true'.

...

By this way, for all *n*,

the given statement is true.

So the given statement is true.

## Example

Rewrite [sum of (4*k* - 1)]

into 4⋅[sum of *k*] - [sum of 1].

Basic properties of summation

As as *k* goes from 1 to *n*,

[sum of *k*] = *n*(*n* + 1)/2,

[sum of 1] = *n* (= 1⋅*n*).

So (given) = 4⋅[*n*(*n* + 1)/2] - *n*.