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- Thread starter carameled
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\(\displaystyle \sum_{i=1}^{n}\left(3i+1\right)=\frac{n}{2}(3n+5)\)

The first thing we want to do is confirm the base case \(P_1\) is true:

\(\displaystyle \sum_{i=1}^{1}\left(3i+1\right)=\frac{1}{2}(3(1)+5)\)

Is this true?

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\(\displaystyle \sum_{i=1}^{n}\left(3i+1\right)=\frac{n}{2}(3n+5)\)

The first thing we want to do is confirm the base case \(P_1\) is true:

\(\displaystyle \sum_{i=1}^{1}\left(3i+1\right)=\frac{1}{2}(3(1)+5)\)

Is this true?

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- #4

You don't understand what an induction hypothesis is, or demonstrating the truth of the base case? These are fundamental to induction. What method have you been taught?Wow, well I'm just asking for the prove with math induction. I don't understand any of that..

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You don't understand what an induction hypothesis is, or demonstrating the truth of the base case? These are fundamental to induction. What method have you been taught?

- Jan 30, 2018

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Well, can you answer the question: is the statement true when n= 1?