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$1(1!) + 2(2!) + 3(3!) + \dots + n(n!) = (n+1)! - 1$

How do we prove this by strong induction?

I know how to do it with weak induction, but how would strong induction work with this problem?

Martin Sleziak
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Nishant Roy
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    Weak induction is merely a special case of strong induction. If you’ve proved it by weak induction, you have automatically proved it by strong induction as well. – Brian M. Scott Apr 29 '16 at 20:25
  • The difference between weak and strong induction is in the inductive hypothesis. You assume more with strong induction (assume that the claim is true for $n=1..k$ instead of just assuming that the claim is true for $n=k$). The two inductions, however, give equivalent theories. – Michael Burr Apr 29 '16 at 20:26
  • There are several post about the same sum: http://math.stackexchange.com/questions/301615/prove-by-mathematical-induction-11-22-cdot-cdot-cdot-nn-n http://math.stackexchange.com/questions/326105/sum-i-1n-i-cdot-i-n1-1-by-induction http://math.stackexchange.com/questions/917367/using-induction-to-prove-that-sum-r-1n-r-cdot-r-n1-1 http://math.stackexchange.com/questions/948757/prove-11-dotsnn-n1-1-using-induction http://math.stackexchange.com/questions/1455778/prove-sum-i-1n-i-cdot-i-n1-1 – Martin Sleziak Jan 26 '17 at 16:53

1 Answers1

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The induction step $$(n+1)(n+1)!+(n+1)!=(n+2)!$$

Rene Schipperus
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