Question

Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation

sk = sk − 1 + 2k, for each integer k ≥ 1
s0 = 3.
satisfies the formula

sn = 3 + n(n + 1) for every integer n ≥ 0.

Proof (by mathematical induction):

Answer 1

To prove that the sequence defined by the recurrence relation satisfies the formula sn = 3 + n(n + 1), we need to show that the base case holds and then prove the inductive step.

Step-by-step explanation:

To prove that the sequence defined by the recurrence relation satisfies the formula , we need to show that the base case holds and then prove the inductive step.

Base case:

When , we have . This matches the formula, so the base case holds.

Inductive step:

Assume that the formula holds for some . We want to show that it holds for .

Using the recurrence relation, we have:

sn+1 = sn + 2(n+1)

Using the induction hypothesis, we can substitute in the expression:

sn+1 = (3 + n(n + 1)) + 2(n+1)

Expanding the expression:

sn+1 = 3 + n(n + 1) + 2n + 2

Combining like terms:

sn+1 = 3 + n(n + 1) + 2(n+1)

sn+1 = 3 + (n+1)((n + 1) + 1)

This matches the formula for , so the inductive step holds. Therefore, the formula holds for all integers .