The recurrence relation T(n) = T(n-1) + 2 is given by, where n > 0 and T(0) = 5.
Question
The recurrence relation T(n) = T(n-1) + 2 is given by, where n > 0 and T(0) = 5.
Solution
The given recurrence relation is a linear homogeneous recurrence relation with constant coefficients. Here's how you can solve it:
Step 1: Identify the base case. Here, T(0) = 5.
Step 2: Start substituting the values from the base case.
For n=1, T(1) = T(1-1) + 2 = T(0) + 2 = 5 + 2 = 7.
Step 3: Continue this process:
For n=2, T(2) = T(2-1) + 2 = T(1) + 2 = 7 + 2 = 9.
For n=3, T(3) = T(3-1) + 2 = T(2) + 2 = 9 + 2 = 11.
And so on.
Step 4: From the above steps, we can see a pattern that T(n) = 2n + 5 for all n ≥ 0.
So, the given recurrence relation T(n) = T(n-1) + 2 can be solved as T(n) = 2n + 5.
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