Question

Determine the sum of the first k odd positive integers for a number of values of k. What generalizations occur to you? Are your inferences correct for all positive integers k?

Answer 1

`S_(n) = \sum_(k=1)^(n) (2k-1) = n^2`

Explanation:

Let's take a look at the first few odd numbers and their sum.

Lets define
`O_(k)` as the
`kth` odd number as:

`O_(k) = 2k-1`

So we have:

`O_(1) = 1\\O_(2) = 3\\O_(3) = 5\\O_(4) = 7\\`

And lets define the sum of all the odd numbers from
`k=1` to
`k=n` as:

`S_(n) = \sum_(k=1)^n O_(k) = \sum_(k=1)^n (2k-1)`

Lets now check some values of said sum:

`S_(1) = 1\\S_(2) = 1 + 3 = 4\\S_(3) = 1 + 3 + 5 = 9\\S_(4) = 1+3+5+7 = 16`

We can then observe than the sum up to
`n` equals
`n^2`

Let us then prove that this is the case by Induction.

First of all, we can prove this by an Induction Proof because we are taking all positive Integers. This is, we are working with the set of natural numbers
`\mathbb{N}`.

We want to prove that

`P(n) = S_(n) = \sum_(k=1)^n = n^2 \forall n\in \mathbb{N}`

This is, we want to prove that the sum of all odd numbers from
`1` to
`n` equals
`n^2` for all natural numbers.

Now, in order to prove something by Induction we need to check 2 things:

`1) The\ base\ case . \ The\ statement\ holds\ for\ n=1\\2) The\ inductive\ step.\ Prove\ that\ if\ the\ statement\ holds\ for\ n\ then\ it\ must\ hold\ for\ n+1\\`

`P(1)` is immediate:

`P(1) = \sum_(k=1)^1 2k-1 = 1 = 1^2`

Now let's assume the statement holds for
`P(n)` and let's take a look at
`P(n+1)`

`P(n+1) = \sum_(k=1)^(n+1) 2k-1`

And we can rewrite it by taking the last term out as:

`P(n+1) = \sum_(k=1)^n 2k-1 \ + 2.(n+1) - 1`

And by inductive hypothesis we know that
`\sum_(k=1)^n 2k-1 = n^2`

and then:

`P(n+1) = \sum_(k=1)^n 2k-1 \ + 2.(n+1) -1  = n^2 + 2n +2 -1 = n^2 +2n +1 = (n+1)^2`

And we have the proof we were looking for!