Let S₀=initial number of sides=3, P₀=initial perimeter=3, L₀=initial side length=1, A₀=initial area=√3/4. In the first generation (n=1), each side becomes 4 sides, so S₁=4×3=12. L₁=1/3 and P₁=S₁L₁=12/3=4.

The area increases by the area of the smaller triangles that grew from the initial number of sides (S₀), one triangle per side. But the length of the sides of the triangle is L₁=1/3, so the added area is S₀(√3/4)L₁²=3(√3/4)/9. So A₁=A₀+S₀(√3/4)L₁²=(√3/4)(1+1/3)=1/√3.

Let’s generalise (using square brackets to denote subscript):

S[n]=4S[n-1]; L[n]=⅓L[n-1]; P[n]=S[n]L[n];

A[n]=A[n-1]+S[n-1](√3/4)L[n]².

We can work out expressions in n only:

S[n]=3(4ⁿ); L[n]=⅓ⁿ; P[n]=3(4ⁿ)(⅓ⁿ)=3(4/3)ⁿ for n≥0.

But for A we need to do something slightly different.

Let a[n] be such that A[n]=A[n-1]+a[n].

Therefore a[n]=S[n-1](√3/4)L[n]²=3(4ⁿ⁻¹)(√3/4)⅓²ⁿ=(4ⁿ⁻²√3)⅓²ⁿ⁻¹ for n>0 and a₀=0.

A[n]=A₀+∑aᵣ for r in [1,n].

∑aᵣ=∑(4ⁿ⁻²√3)⅓²ⁿ⁻¹ for n>0 = √3/12+√3/27+4√3/243+16√3/2187+...

∑aᵣ=(√3/12)(1+(4/9)+(4/9)²+(4/9)³+...+(4/9)ⁿ⁻¹)=

(√3/12)(1-(4/9)ⁿ)/(1-4/9)=

We have a GP in the brackets which sums to (1-(4/9)ⁿ)/(1-4/9)=(9/5)(1-(4/9)ⁿ).

So ∑aᵣ=(3√3/20)(1-(4/9)ⁿ) and

A[n]=√3/4+(3√3/20)(1-(4/9)ⁿ)=(1+(3/5)(1-(4/9)ⁿ)(√3/4) for n≥0.

SOLUTIONS

a. Number of sides, S[n]=3(4ⁿ)

b. Area, A[n]=(1+(3/5)(1-(4/9)ⁿ)(√3/4)

c. Perimeter, P[n]=3(4/3)ⁿ