Look how the pyramid is constructed.

The pyramid is constructed inside an equilateral triangle of size n sticks. It is filled with small equilateral triangles of size 1.

If `a_n ` is the number of small triangles in the big triangle of side n.

`b_n` is the number of sticks needed to fill the big triangle of side n.

It build the next step, we just need to add n+1 small upward traingles of side 1

/_\ /_\ ... /_\

Which means we add 3(n+1) sticks.

Therefore `b_(n+1)=b_n+3(n+1)`

How many small triangles were added?

(n+1) upward and n downward. i.e 2n+1 in total.

`a_(n+1)=2n+1+a_n=2(n+1)-1+a_n`

Let's try to find a expression of `b_n` as a function of n

`b_n=3+3*2+3*3+3*4+....+3*n=3(1+2+...+n)=3*n(n+1)/2`

b_n=3n(n+1)/2

Let's find `a_n`

`a_n=2*1-1+2*2-1+2*3-1+...2*n-1=2(1+2+3+...+n)-1-1-...-1`

`a_n=2n(n+1)/2-n=n(n+1)-n=n^2.`

**Therefore, at the level n there are** `3n(n+1)/2` **sticks and** `n^2 ` **triangles.**

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