Notes PDF
Introductory Problem
1. The whole shape pictured at the right has base and altitude 1. Find the area of the shaded red part remaining after the indicated squares are removed.
Answer
1. The area is 2/3.
Solution
1. The removed squares have side length 1/2, 1/4, 1/8, …, so the removed areas are: (1/2)2 + (1/4)2 + (1/8)2 + … = (1/2)2 + (1/22)2 + (1/23)2 + … = (1/22) + (1/22)2 + (1/22)3 + … = 1/4 + (1/4)2 + (1/4)3 +…This geometric series sums to (1/4)/(1 – (1/4)) = 1/3. Then the area of the shaded region is 1 – 1/3 = 2/3.
2. The whole shape pictured at the right has base and altitude 1. Find the area of the shaded red part remaining after the indicated triangles are removed.
Answer
1. The area of the triangles removed sums to 1/2, so the area remaining is 0.
Solution
1. We see 1 removed triangle with base = altitude = 1/2, three with base = altitude = 1/4, 9 with base = altitude = 1/8, and so on. The removed areas sum to (1/2)(1/2)(1/2) + 3(1/2)(1/4)(1/4) + 9(1/2)(1/8)(1/8) + …= (1/2)(1/4 + 3/42 + 9/43 + …)= (1/8)(1 + 3/4 + (3/4)2 + …) The geometric series sums to 1/(1 – 3/4) = 4, so the areas of the removed triangles sum to 1/2.The area of a triangle with base 1 and altitude 1 is 1/2, so the shaded region has area = 0.
3. Express the repeating decimal 1.1717171717… as a rational number.
Answer
1. The rational form of the repeating decimal 1.1717171717… is 116/99.
Solution
1. Write the repeating decimal 1.1717171717… as 1 + 17/100 + 17/1002 + 17/1003 + … = 1 + (17/100)(1 + 1/100 + 1/1002 + … = 1 + (17/100)(1/(1 – 1/100)) = 1 + 17/99 = 116/99