The recurrence T(n) = 2T(n-1) + n, for n≥2 and T(1) = 1 evaluates to
Question
The recurrence T(n) = 2T(n-1) + n, for n≥2 and T(1) = 1 evaluates to
Solution 1
To solve the recurrence relation T(n) = 2T(n-1) + n, we can use the method of iteration.
Step 1: Start with the given recurrence relation: T(n) = 2T(n-1) + n
Step 2: Substitute n-1 into the recurrence relation: T(n-1) = 2T(n-2) + n-1
Step 3: Substitute T(n-1) from step 2 into the original recurrence relation: T(n) = 2[2T(n-2) + n-1] + n = 4T(n-2) + 2n - 2 + n = 4T(n-2) + 3n - 2
Step 4: Repeat the process for T(n-2): T(n-2) = 2T(n-3) + n-2
Step 5: Substitute T(n-2) from step 4 into the equation from step 3: T(n) = 4[2T(n-3) + n-2] + 3n - 2 = 8T(n-3) + 4n - 8 + 3n - 2 = 8T(n-3) + 7n - 10
If we continue this process, we can see a pattern emerging. The general form after k steps is:
T(n) = 2^k * T(n-k) + k*n - (2^k - 1)
When n=k, T(n) = 2^n * T(0) + n^2 - (2^n - 1). Given that T(1) = 1, we can say that T(0) = 0. So,
T(n) = n^2 - (2^n - 1)
This is the closed form solution for the given recurrence relation.
Solution 2
To solve the recurrence relation T(n) = 2T(n-1) + n, we can use the method of iteration.
Step 1: Start with the given recurrence relation: T(n) = 2T(n-1) + n
Step 2: Substitute T(n-1) from the recurrence relation: T(n) = 2[2T(n-2) + (n-1)] + n = 4T(n-2) + 2(n-1) + n
Step 3: Substitute T(n-2) from the recurrence relation: T(n) = 4[2T(n-3) + (n-2)] + 2(n-1) + n = 8T(n-3) + 4(n-2) + 2(n-1) + n
If we continue this process, we can see a pattern emerging. After k steps, the equation will look like this:
T(n) = 2^kT(n-k) + k(n-k+1)
Step 4: We continue this process until we reach the base case, which is T(1) = 1. This happens when n-k = 1, or k = n-1. Substituting k = n-1 into the equation from Step 3 gives:
T(n) = 2^(n-1)T(1) + (n
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