1964 AMC 12 Problem 31

Attempt Problem 31 of the 1964 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1964 AMC 12 solutions, or check the answer key.

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31.

Let

\begin{aligned} f(n) &=\frac{5+3\sqrt5}{10} \left(\frac{1+\sqrt5}{2}\right)^n\\ &\quad+\frac{5-3\sqrt5}{10} \left(\frac{1-\sqrt5}{2}\right)^n. \end{aligned}

Then

f(n+1)-f(n-1),

expressed in terms of

f(n),

equals:

\dfrac12f(n)

f(n)

2f(n)+1

f^2(n)

\dfrac12\left(f^2(n)-1\right)

Answer: B

Concepts:recursion Fibonacci exponent

Difficulty rating: 1900

Small hint:

Call the two exponential bases

\alpha

and

\beta

Big hint:

Both roots of

r^2-r-1=0

satisfy

r^{n+1}-r^{n-1}=r^n

Solution:

Each base

r=\frac{1\pm\sqrt5}{2}

satisfies

r^2=r+1,

so

r^2-1=r.

Therefore

r^{n+1}-r^{n-1}=r^{n-1}(r^2-1)=r^n.

Applying this identity term by term to the defining linear combination gives

f(n+1)-f(n-1)=f(n).

Thus, the correct answer is B.