Consider the following recurrence relation. p(n) = {1 if n = 0 p(n-1) + n^2 if n

Question

Consider the following recurrence relation. p(n) = {1 if n = 0 p(n-1) + n^2 if n > 0 Compute the first eight values of P(n). (Enter your answers as a comma-separated list of values ordered with respect to increasing n.) Analyze the sequences of differences. (Enter your answers as a comma-separated list of values ordered with respect to increasing n.) first differences: second differences: third differences: What does this suggest about the closed form solution? This suggests that the closed form will have degree Find a good candidate for a closed-form solution. Prove that your candidate solution is the correct closed-form solution. (Induction on n.) Let f(n) = Base Case: If n = 0, the recurrence relation says that P(0) = 1, and the formula says that f(0) =, so they match. Inductive Hypothesis: Suppose as inductive hypothesis that P(k - 1) = for some k > 0. Inductive Step: Using the recurrence relation, P(k) = p(k - 1) + k^2, by the second part of the recurrence relation + k2, by inductive hypothesis so, by induction, P(n) = f(n) for all n >= 0.

Explanation / Answer

(a) P0 = 1

P(1) = P(0) + 12 = 2

P(2) = P(1) + 22 = 6

P(3) = P(2) + 32 = 15

P(4) = P(3) + 42 = 31

P(5) = P(4) + 52 = 56

P(6) = P(5) + 62 = 92

P(7) = P(6) + 72 = 141

P(8) = P(7) + 82 = 205

(b) First differences: 4, 9, 16, 25, 36, 49, 64

Second differences: 5, 7, 9, 11, 13, 15

Third differences: 2, 2, 2, 2, 2

This suggests that the closed form solution will have degree 3

(c) P(n) = P(n-1) + n2

P(n) = P(n-2) + (n-1)2  + n2

P(n) = P(0) + 12 + 22  ....(n-1)2  + n2

P(n) = 1 + n(n+1)(2n+1)/6

(d) f(n) = 1 + n(n+1)(2n+1)/6

f(0) = 1

P(k-1) = 1 + (k-1)k(2k-1)/6

P(k) = P(k-1) + k2

P(k) = 1 + (k-1)k(2k-1)/6 + k2

P(k) = 1 + k(k+1)(2k+1)/6

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