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In the book Discrete Mathematics and Its Applications, 8e, Kenneth Rosen it is mentioned that:

... mathematical induction and strong induction are equivalent. That is, each can be shown to be a valid proof technique assuming that the other is valid.

I interpret this to mean that neither proof technique can be used where the other could not be used. That being the case, why bother about strong induction at all?

Sandeep
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If you are trying to prove F(n) is true, then it is more convenient to work with the fact that for all 1 <= k < n, F(k) is true, rather than just k = n-1 because you may need to depend on a sub-problem other than F(n-1).

For example, if we wanted to prove the Fibonacci recurrence T(n) = T(n-1) + T(n-2) has some closed form g(n), then we could leverage strong induction for convenience.

Scott Hahn
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