Walkabout wrote:
In a certain sequence, the term \(x_n\) is given by the formula \(x_n=2*x_{n-1}-\frac{1}{2}*x_{n-2}\) for all \(n\geq{2}\). If \(x_0=3\) and \(x_1=2\), what is the value of \(x_3\)?
(A) 2.5
(B) 3.125
(C) 4
(D) 5
(E) 6.75
(Note that the * symbol in both the question stem and in this solution indicates multiplication.)
We are given that X(n) = 2 * X(n-1) – ½ * X(n-2), for all n=>2. This is called a recursive formula, which means that we need to know prior terms before we can compute the subsequent terms. For example, if we want to know X(2), we must know both X(1) and X(0), because X(2) is equal to 2 * X(1) – ½ * X(0).
We are given X(1) = 2 and X(0) = 3. So, when n is 2, X(2) would be calculated as follows:
X(2) = 2 * X(1) – ½ * X(0)
X(2) = 2 * 2 – ½ * 3
X(2) = 4 – 1.5
X(2) = 2.5
Now we are ready to determine the value of X(3). In this case, n = 3, X(1) is 2, and X(2) is 2.5. We plug these values into the recursive formula given in the question stem:
X(3) = 2 * X(2) – ½ * X(1)
X(3) = 2 * 2.5 – ½ * 2
X(3) = 5 – 1
X(3) = 4
Answer is C.
I approached this problem by brute force if you will in that I started by plugging in 3 to X (so x sub 3) to see what I was missing. Is this a bad approach to use when solving other recursive formulas? Thank you