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  • 5 years ago

is any one good at doing proofs by induction?

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  1. anonymous
    • 5 years ago
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    example: for all numbers n = 0,1,2,...., the sum 0+1+...+n is given by \[\frac{n(n+1)}{2}\] First we do the base case, which means to verify that the statement is true for n = 0. Indeed, if we plug in n = 0, we get that 0 = 0(1)/2 = 0. Next we have the induction step, where we assume that the statement is true for n, and prove it for n+1: Assume that \[0+1+...+n =\frac{n(n+1)}{2}\] Then by adding n+1 at the end we get: \[0+1+...+n+(n+1) = \frac{n(n+1)}{2}+(n+1) \] which when simplified gives \[\frac{(n+1)(n+2)}{2}\] and we've shown that the statement hold for n+1. Now, the base case has shown that n = 0 satisfies the statement, and the inductive step shows that since n = 0 holds, so does n = 1, and thus n = 2, etc... we've showed "by induction" that it holds for every n.

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