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Given $a_0 = 1$ and $a_1 = 5,$ and the general relation \[a_n^2 - a_{n - 1} a_{n + 1} = (-1)^n\]for $n \ge 1,$ find $a_3.$

 Feb 23, 2023
 #1
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Hard problem!

 

We can use the given relation to find a_3, as follows:

a_3^2 - a_2 a_4 = (-1)^3 (using n = 3 in the given relation)

We know a_0 and a_1, and we can find a_2 using the relation with n = 2:

a_2^2 - a_1 a_3 = (-1)^2

Substituting the given values a_0 = 1 and a_1 = 5, we get:

a_2^2 - 5a_3 = 1 (equation 1)

We need to find a_3, so we also need to find a_4. We can use the given relation again with n = 4:

a_4^2 - a_3 a_5 = (-1)^4

Substituting a_0 = 1 and a_1 = 5, and using equation 1 to eliminate a_5 in favor of a_3, we get:

a_4^2 - a_3 (a_2 + 1/a_3) = 1

Simplifying, we get:

a_4^2 - a_2 a_3 - 1 = 0 (equation 2)

Now we have two equations with two unknowns, a_2 and a_3. We can solve for a_2 in equation 1 and substitute in equation 2, to get a quadratic equation in a_3:

(a_3^2 - 1/5)^2 - 5a_3^2 - 1 = 0

Expanding and simplifying, we get:

a_3^4 - (5/2)a_3^2 + (24/25) = 0

This is a quadratic equation in a_3^2, which we can solve using the quadratic formula:

a_3^2 = [(5/2) +/- sqrt((5/2)^2 - 4*(24/25))]/2

a_3^2 = [5 +/- sqrt(5/4)]/2

Taking the positive root, we get:

a_3^2 = (5 + sqrt(5))/4

Taking the square root, we finally get:

a_3 = sqrt((5 + sqrt(5))/4)

Therefore, a_3 is approximately 1.618, which is the golden ratio.

 Feb 23, 2023
 #2
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The expression an^2 - a(n - 1) a(n + 1) = (-1)^n can be rearranged to give an=(-1)^n + a(n - 1) a(n + 1). When n = 3, this gives us a3=(-1)^3 + a2 a4. Since a2 = 5 and a4 is unknown, we can substitute a2 = 5 and solve for a4. This gives us a4=5/a3. Substituting this expression into the equation for a3 gives us a3=(-1)^3 + 5/a3. Solving for a3 gives us a3=2.

 Feb 23, 2023

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