A sequence {ai}is defined by the recurrence relation an=40−4an−1 with a0=−4. There exists real valued constants r,s and t such that ai=r⋅si+t for all non-negative integers i. Determine r2+s2+t2.
To determine the real-valued constants r, s, and t, we need to find a closed form expression for the given recurrence relation. Let's start by finding the first few terms of the sequence:
a0 = -4
a1 = 40 - 4(a0) = 40 - 4(-4) = 56
a2 = 40 - 4(a1) = 40 - 4(56) = -144
a3 = 40 - 4(a2) = 40 - 4(-144) = 616
We can observe a pattern here. Let's write out a few more terms:
a4 = 40 - 4(a3) = 40 - 4(616) = -2344
a5 = 40 - 4(a4) = 40 - 4(-2344) = 9552
a6 = 40 - 4(a5) = 40 - 4(9552) = -38168
It seems that the terms are alternating between positive and negative values and increasing in magnitude. This suggests that we can express the terms as a linear combination of two sequences, one increasing and one decreasing.
Let's define two sequences: {b_i} and {c_i}.
b_0 = 1, c_0 = 0
b_i = 4b_{i-1}
c_i = -4c_{i-1} + 40
Now let's express a_i in terms of b_i and c_i:
a_i = b_i + c_i
We can find recursive expressions for b_i and c_i:
b_i = 4b_{i-1}
c_i = -4c_{i-1} + 40
To obtain a closed-form expression for b_i and c_i, we can solve these recursive relations.
Let's find a closed-form expression for b_i:
b_i = 4b_{i-1}
b_0 = 1
By iteratively substituting, we have:
b_i = 4^i * b_0 = 4^i
Next, let's find a closed-form expression for c_i:
c_i = -4c_{i-1} + 40
c_0 = 0
By iteratively substituting, we have:
c_i = (-4)^i * c_0 + 10(1 - (-4)^i) = 10 - 14^i
Now, let's substitute the closed-form expressions of b_i and c_i into a_i:
a_i = b_i + c_i = 4^i + 10 - 14^i
Now we can express ai=r⋅si+t:
a_i = r * s^i + t
4^i + 10 - 14^i = r * s^i + t
Comparing the coefficients, we have:
r = 1
s = 4
t = 10
Thus, r^2 + s^2 + t^2 = 1^2 + 4^2 + 10^2 = 1 + 16 + 100 = 117.