On a straight line ℓ, we have an infinite sequence of circles Γn, each with radius 1/2^n, such that Γn is externally tangential to the circles Γn−1,Γn+1 and the line ℓ. Consider another infinite sequence of circles Cn, each with radius rn, such that Cn is externally tangential to Γn,Γn+1 and ℓ. The expression ∑i=1 to ∞ ri can be expressed as a−√b, where a and b are positive integers. What is the value of a+b?
Clarification: In this problem, we have a row of circles placed on a line. All points of tangency are distinct. The circle Cn is uniquely determined
okay at least somebody give a hint
To find the value of a+b, we need to determine the expression ∑i=1 to ∞ ri. Let's break down the problem step by step.
First, let's find the radius of the circle Cn in terms of n. As given in the problem, Cn is externally tangential to Γn and Γn+1. Since the radius of Γn is 1/2^n, the radius of Cn will be the sum of the radii of Γn and Γn+1.
Therefore, the radius of Cn, rn, is given by rn = (1/2^n) + (1/2^(n+1)).
Now, let's simplify the expression for rn. We can rewrite rn as rn = (1/2^n) + (1/2^n * 1/2). Simplifying further, rn = (2/2^n) + (1/2^(n+1)). Combining the terms under a common denominator, rn = (3/2^n) * (1/2).
Hence, the expression for rn is rn = 3/2^(n+1).
Now, let's calculate the sum of the infinite sequence ∑i=1 to ∞ ri. We substitute the expression for rn into the summation:
∑i=1 to ∞ ri = ∑i=1 to ∞ (3/2^(i+1))
We can rewrite the above expression as ∑i=1 to ∞ (3/2^2 * 1/2^i).
Further simplifying, we have ∑i=1 to ∞ (3/4 * 1/2^i).
Using the formula for the sum of an infinite geometric series, we get:
∑i=1 to ∞ (3/4 * 1/2^i) = (3/4) * (1 / (1 - 1/2)) = (3/4) * (1/ (1/2)) = (3/4) * 2 = 3/2.
Therefore, the sum ∑i=1 to ∞ ri is equal to 3/2.
Finally, we need to express the answer as a - √b, where a and b are positive integers. In this case, a = 3 and b = 4.
Therefore, the value of a + b = 3 + 4 = 7.
Hence, the answer is 7.