Question

What is the formula for the `n`th term of the sequence `1, 3, 6, 10, 15,...` ?

Answer 1

`a_n = 1/2n(n+1)`

Explanation:

These are triangular numbers:

`0color(white)(0000)0color(white)(000000)0color(white)(00000000)0color(white)(0000000000)0`
`color(white)(0000)0color(white)(0)0color(white)(0000)0color(white)(0)0color(white)(000000)0color(white)(0)0color(white)(00000000)0color(white)(0)0`
`color(white)(0000000000)0color(white)(0)0color(white)(0)0color(white)(0000)0color(white)(0)0color(white)(0)0color(white)(000000)0color(white)(0)0color(white)(0)0`
`color(white)(000000000000000000)0color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0000)0color(white)(0)0color(white)(0)0color(white)(0)0`
`color(white)(0000000000000000000000000000)0color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0)0`

Geometrically, you can see that such a triangle is one half of a parallelogram with base `n+1` and height `n`, e.g. `a_5:`

`color(white)(0000)0color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)`
`color(white)(000)0color(white)(0)0color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)`
`color(white)(00)0color(white)(0)0color(white)(0)0color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)`
`color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)`
`0color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0)color(blue)(0)`

Such a parallelogram has a total count of `n(n+1)`, so the triangle is half of that.

In other words:

`a_n = sum_(k=1)^n k = 1/2n(n+1)`

Answer 2

`(n^2+n)/2`

Explanation:

Starting with 1 you can begin to see a pattern develop....

`a_1=1`

`a_2=a_1+2=1+2=3`

`a_3=a_2+3=3+3=6`

`a_4=a_3+4=6+4=10`

`a_5=a_4+5=10+5=15`

`vdots`

`a_n=a_(n-1)+n`

Since every term is merely the sum of the previous term plus the next counting number we have.

`a_n=sum_(k=1)^nk=(n(n+1))/2`

To test this we use the first formula for `a_(n+1)`

`a_(n+1)=a_n+(n+1)`

`=> a_(n+1)=(n(n+1))/2+(n+1)=(n^2+n)/2+(2n+2)/2=(n^2+3n+2)/2`

by the hypothesis

`a_(n+1)=((n+1)(n+2))/2=(n^2+3n+2)/2`

They are equal...

So the hypothesis is proven

This is an example of "Weak" Induction