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 September 29th, 2017, 10:49 PM #1 Newbie   Joined: Sep 2017 From: czech republic Posts: 2 Thanks: 0 how to prove this without mathematical induction Hello everyone. a) Σ (from k=1 to n) (k^2+1)*k!=n*(n+1)! b) Σ (from k=1 to n) 2^(n-k)*k*(k+1)!=(n+2)!-2^(n+1) I know how to solve it with mathematical induction. However, I can't. I have to find a different way. But I don't see how. Please, could you tell me how to do it? Thank you. Last edited by skipjack; September 29th, 2017 at 11:00 PM.
 September 29th, 2017, 11:39 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,739 Thanks: 1810 (a) If f(k) ≡ k(k + 1)!, (k² + 1)k! = f(k) - f(k - 1) and f(0) = 0, so the sum is f(1) - f(0) + f(2) - f(1) + ... + f(n) - f(n - 1), which is f(n). Although the above isn't set out as a use of mathematical induction, it's equivalent to the use of mathematical induction. The part (b) of your question can be tackled similarly.
 September 30th, 2017, 11:28 AM #3 Newbie   Joined: Sep 2017 From: czech republic Posts: 2 Thanks: 0 Thank you, now I understand. However, I don't know how to solve second equation, when the n is on both sides. It is different. How can I do it?

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