## Tuesday, April 4, 2006

### Real Positive. Topic: Algebra/Inequalities/Sequences & Series. Level: Olympiad.

Problem: (Math Olympiad Treasures - 1.3.13) Suppose the sequence $a_1, a_2, \ldots, a_n$ satisfies the following conditions:

$a_1 = 0$,$|a_2| = |a_1+1|$, $\ldots$, $|a_n| = |a_{n-1}+1|$.

Prove that

$\frac{a_1+a_2+\cdots+a_n}{n} \ge -\frac{1}{2}$.

Solution: Consider squaring the relations given in the sequence to obtain

$a_2^2 = (a_1+1)^2 = a_1^2+2a_1+1$

$a_3^2 = (a_2+1)^2 = a_2^2+2a_2+1$

$\ldots$

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

$a_2^2+a_3^2+\cdots+a_{n+1}^2 = a_1^2+a_2^2+\cdots+a_n^2+2(a_1+a_2+\cdots+a_n)+n$, or

$a_{n+1}^2 = 2(a_1+a_2+\cdots+a_n)+n \ge 0$.

Then we have

$\frac{a_1+a_2+\cdots+a_n}{n} \ge -\frac{1}{2}$,

as desired. QED.

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Comment: Absolute values are easy to get rid of by squaring things, usually giving us nice results. This method can be used in a variety of problems, but many times with sequences recursively defined we see it at its best. The following AIME problem can be solved this way.

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Practice Problem: (2006 AIME I - #15) Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1$, find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|$.