1958 AMC 12 Problem 40

Attempt Problem 40 of the 1958 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 1958 AMC 12 solutions, or check the answer key.

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40.

Given

a_0=1,

a_1=3,

and the general relation

a_n^2-a_{n-1}a_{n+1}=(-1)^n

for

n\ge1.

Then

a_3

equals:

\dfrac{13}{27}

33

21

10

-17

Answer: B

Concepts:recursion substitution

Difficulty rating: 1210

Small hint:

Use the recurrence first with

n=1

to find

a_2

Big hint:

Then use it with

n=2

to solve for

a_3

Solution:

For

n=1,

3^2-(1)a_2=-1,

so

a_2=10.

For

n=2,

10^2-3a_3=1,

which gives

a_3=33.

Thus, the correct answer is B.

Problem 40 in other years