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Difference between revisions of "1970 AHSME Problems/Problem 16"

m (See also)
(Solution)
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\text{(E) } 26</math>
 
\text{(E) } 26</math>
  
== Solution ==
+
== Solution =
<math>\fbox{C}</math>
+
We can chug through the recursion to find the answer is <math>\fbox{C}</math>.
 +
 
 +
==Sidenote==
 +
All the numbers in the sequence <math>F(n)</math> are integers. In fact, the function <math>F</math> satisfies <math>F(n)=4F(n-2)-F(n-4)</math>. (Prove it!).
  
 
== See also ==
 
== See also ==

Revision as of 00:01, 25 May 2015

Problem

If $F(n)$ is a function such that $F(1)=F(2)=F(3)=1$, and such that $F(n+1)= \frac{F(n)\cdot F(n-1)+1}{F(n-2)}$ for $n\ge 3,$ then $F(6)=$

$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 7\quad \text{(D) } 11\quad \text{(E) } 26$

= Solution

We can chug through the recursion to find the answer is $\fbox{C}$.

Sidenote

All the numbers in the sequence $F(n)$ are integers. In fact, the function $F$ satisfies $F(n)=4F(n-2)-F(n-4)$. (Prove it!).

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
All AHSME Problems and Solutions

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