Difference between revisions of "2008 Mock ARML 2 Problems/Problem 8"

Revision as of 08:54, 30 May 2008

Given that $\sum_{i = 0}^{n}a_ia_{n - i} = 1$ and $a_n > 0$ for all non-negative integers $n$ , evaluate $\sum_{j = 0}^{\infty}\frac {a_j}{2^j}$ .

$a_0^2=1\Rightarrow a_0=1$

$a_0a_1+a_1a_0=1\Rightarrow a_1=\dfrac{1}{2}$

$a_0a_2+a_1^2+a_2a_0=1\Rightarrow a_2=\dfrac{3}{8}$

$a_0a_3+a_1a_2+a_2a_1+a_3a_0=2a_3+\dfrac{3}{8}=1\Rightarrow a_3=\dfrac{5}{16}$

We make a conjecture that $a_i=\dfrac{F_{i+3}}{2^{i+1}}$ , where $F_n$ is the $n$ th Fibonacci number and we prove that it is so:

$a_ia_{n-i}=\dfrac{F_{i+3}\cdot F_{n-i+3}}{2^{n+2}}$

We must prove that $\sum_{i = 0}^{n}\dfrac{F_{i+3}\cdot F_{n-i+3}}{2^{n+2}} = 1$ .

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Revision as of 08:53, 30 May 2008 (view source) 1=2 (talk \| contribs) (New page: ==Problem== Given that <math>\sum_{i = 0}^{n}a_ia_{n - i} = 1</math> and <math>a_n > 0</math> for all non-negative integers <math>n</math>, evaluate <math>\sum_{j = 0}^{\infty}\frac {a_j}{...)		Revision as of 08:54, 30 May 2008 (view source) 1=2 (talk \| contribs) (→‎See also) Newer edit →
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	==See also==		==See also==
		+	{{Mock ARML box\|year = 2008\|n = 2\|num-b=7\|after=Final Question\|source = 206880}}