Problem: An equilateral triangle is drawn with a side of length a. A new equilateral triangle is formed by joining the mid-points of the sides of the first one. Then a third equilateral triangle is formed by joining the mid-points of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is:
Answer Choices:
A. Infinite
B. 541​a
C. 2a
D. 6a
E. 421​a
Solution:
Let Pk​ be the perimeter of the k th triangle.
Then Pk+1​=21​Pk​.
S​=P1​+P2​+P3​+⋯=3a+21​⋅3a+21​⋅23​a+⋯=3a(1+21​+41​+⋯)=3a1−(1/2)1​=6a​
