Carcass wrote:
\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)
A. \(\frac{1}{2^7}\)
B. \(\frac{1}{2^8}\)
C. \(\frac{1}{2^9}\)
D. \(\frac{1}{2^{13}}\)
E. \(\frac{1}{2^{45}}\)
Key property: \(\frac{1}{x^n} = x^{-n}\)For example, \(\frac{1}{5^8} = 5^{-8}\)
So, we can rewrite our sum as follows: \(2^{-10} + 2^{-11} + 2^{-12} + 2^{-12}\)
Factor out \(2^{-12}\) to get: \(2^{-12}(2^2 + 2^1 + 1 + 1)\)
Evaluate to get: \(2^{-12}(4 + 2 + 1 + 1)\)
Simplify: \(2^{-12}(8)\)
Rewrite as follows: \((2^{-12})(2^3)\)
Apply the product law to get: \(2^{-9}\)
Rewrite as: \(\frac{1}{2^{9}}\)
Answer: C
Cheers,
Brent