Prove that \(\sum_{k=1}^{n} (3k - 1) = \frac{n(3n + 1)}{2}\) for all \(n \in \mathbb{Z}^+\) using mathematical induction by completing the steps of the proof as indicated. (a) Initial statement: (1 mark) (b) Inductive step: Assume the result is true for \(n = m\) and prove it is true for \(n = m + 1\). (2 marks)
Specialist Mathematics ยท Unit 3 ยท Mathematical induction and trigonometric proofs ยท Mathematical induction
Prove results for sums for any positive integer ๐.
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Use mathematical induction to prove that \(\displaystyle\sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(n+2)}{3}\) for \(n \in \mathbb{Z}^+\).
Use the stimulus to prove that \(\displaystyle\sum_{k=1}^{n} (3k - 1) = \frac{n(3n + 1)}{2}\) for all \(n \in \mathbb{Z}^+\).
When using proof by mathematical induction to show that \(\displaystyle\sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(n+2)}{3}\) for all \(n \in \mathbb{Z}^+\), the inductive step requires proving
A researcher records the total number of bacteria in a culture at the end of each hour. The table below shows the cumulative count for the first 4 hours. Use mathematical induction to prove that if the cumulative count after \(n\) hours is given by \(\displaystyle\sum_{k=1}^{n} k(k+1)\), then \(\displaystyle\sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(n+2)}{3}\) for all \(n \in \mathbb{Z}^+\).