Categories
AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and Integers | AIME I, 2007 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.

Sequence and Integers – AIME I, 2007


A sequence is defined over non negetive integral indexes in the following way \(a_0=a_1=3\), \( a_{n+1}a_{n-1}=a_n^{2}+2007\), find the greatest integer that does not exceed \(\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}\)

  • is 107
  • is 224
  • is 840
  • cannot be determined from the given information

Key Concepts


Sequence

Inequalities

Integers

Check the Answer


But try the problem first…

Answer: is 224.

Source
Suggested Reading

AIME I, 2007, Question 14

Elementary Number Theory by David Burton

Try with Hints


First hint

\(a_{n+1}a_{n-1}\)=\(a_{n}^{2}+2007\) then \(a_{n-1}^{2} +2007 =a_{n}a_{n-2}\) adding these \(\frac{a_{n-1}+a_{n+1}}{a_{n}}\)=\(\frac{a_{n}+a_{n-2}}{a_{n-1}}\), let \(b_{j}\)=\(\frac{a_{j}}{a_{j-1}}\) then \(b_{n+1} + \frac{1}{b_{n}}\)=\(b_{n}+\frac{1}{b_{n-1}}\) then \(b_{2007} + \frac{1}{b_{2006}}\)=\(b_{3}+\frac{1}{b_{2}}\)=225

Second Hint

here \(\frac{a_{2007}a_{2005}}{a_{2006}a_{2005}}\)=\(\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}\) then \(b_{2007}\)=\(\frac{a_{2007}}{a_{2006}}\)=\(\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}\)\( \gt \)\(\frac{a_{2006}}{a_{2005}}\)=\(b_{2006}\)

Final Step

then \(b_{2007}+\frac{1}{b_{2007}} \lt b_{2007}+\frac{1}{b_{2006}}\)=225 which is small less such that all \(b_{j}\) s are greater than 1 then \(\frac{a_{2006}^{2}+ a_{2007}^{2}}{a_{2006}a_{2007}}\)=\(b_{2007}+\frac{1}{b_{2007}}\)=224.

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.