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Proof Jumble 4/25/22, \(= \frac{N(N+1)+2(N+1)}{2} \), Therefore, the sum…
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Therefore, the sum of the first \(n\) positive integers is given by \(\frac{n(n+1)}{2}\). \(\blacksquare\)
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Accordingly, the inductive hypothesis holds for the next step.
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As the inductive hypothesis has been shown to start and then continue indefinitely, the inductive hypothesis holds for all positive integer values of \(n\).
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We will use this inductive hypothesis to show that the first \(N+1\) positive integers sum to \(\frac{(N+1)((N+1)+1)}{2}\).
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Hypothesize for induction that the sum of the first \(N\) positive integers can be expressed as \(\frac{N(N+1)}{2}\).
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Consider the base case, \(n=1\):
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Pf. We will show the sum of the first \(n\) positive integers is given by \(\frac{n(n+1)}{2}\) by induction on \(n\).
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The expression \(\frac{1(1+1)}{2}\) simplifies to 1 and matches the sum of the first one positive integer as desired.
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