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2021 Math Kangaroo

Problem 23

Problem 23 · 2021 Math Kangaroo Stretch
Algebra & Patterns substitution

The function f is such that \(f(x+y) = f(x) \cdot f(y)\) and \(f(1) = 2\). What is the value of \(\dfrac{f(2)}{f(1)} + \dfrac{f(3)}{f(2)} + \cdots + \dfrac{f(2021)}{f(2020)}\)?

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Answer: E — none of the previous
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Hint 1 of 2
The rule f(x+y)=f(x)f(y) with f(1)=2 forces a familiar formula for f.
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Hint 2 of 2
Simplify a typical term f(n+1)/f(n) before adding.
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Approach: identify the exponential and collapse each term
  1. From f(x+y)=f(x)f(y) and f(1)=2 we get f(n)=2ⁿ.
  2. Each term f(n+1)/f(n) = 2ⁿ⁺¹/2ⁿ = 2, and there are 2020 terms (from f(2)/f(1) to f(2021)/f(2020)).
  3. The sum is 2020 × 2 = 4040, which is not among A–D, so none of the previous.
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