2. Prove by induction that for even positive integer $n$, there exists two odd positive integers such that they sum to $n$. 3. Set $b_n = \left(\sum_{k=1}^n k\right) - n$. Prove by induction: $\sum_{i=1}^n b_i = \frac{(n-1)(n+1)n}{6}$. For this assignment, you may use a result proved in class.
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Base case: For n = 2, we can choose 1 and 1 as the two odd positive integers that sum to 2. Inductive step: Assume that for some even positive integer k, there exist two odd positive integers that sum to k. We want to show that there exist two odd positive Show more…
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