Let ƒÆ=sin −1 7 25 . Consider the sequence of values defined by a n =sin(nƒÆ) . They satisfy the recurrence relation
a n+2 =k 1 a n+1 +k 0 a n ,n�¸N
for some (fixed) real numbers k 1 ,k 0 . The sum k 1 +k 0 can be written as p q , where p and q are positive coprime integers. What is the value of p+q ?
To find the value of p + q, we need to determine the values of k1 and k0 first.
Given that a_n = sin(nĮ), we can rewrite the recurrence relation as:
a_(n+2) = k1 * a_(n+1) + k0 * a_n
Now, plug in n = 0:
a_2 = k1 * a_1 + k0 * a_0
Since a_0 = sin(0Į) = sin(0) = 0 and a_1 = sin(1Į), the equation becomes:
a_2 = k1 * sin(1Į)
Next, plug in n = 1:
a_3 = k1 * a_2 + k0 * a_1
Using the previous equation for a_2, we have:
a_3 = k1 * k1 * sin(1Į) + k0 * sin(1Į)
We can simplify this equation as:
a_3 = (k1^2 + k0) * sin(1Į)
Now, since the sequence defined by a_n = sin(nĮ) satisfies the recurrence relation, we can substitute the values of a_2 and a_3 from the above equations:
k1 * sin(1Į) = (k1^2 + k0) * sin(1Į)
Dividing both sides by sin(1Į), we get:
k1 = k1^2 + k0
Now, we can solve this quadratic equation to find the possible values of k1. Subtracting k1^2 from both sides, we have:
k0 = -k1^2 + k1
So, we have two equations:
k1 = k1^2 + k0
k0 = -k1^2 + k1
Now, we need to find the sum k1 + k0. Adding the two equations, we get:
k1 + k0 = k1^2 + k0 - k1^2 + k1
= 2k1
Therefore, the sum k1 + k0 is equal to 2k1.
Since we do not have the specific values of k1 and k0, we cannot determine the exact value of p + q.