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Mathematics Inclined Week #1: Difficult Category

Problem:
Prove, by mathematical induction, that:


1(1!) + 2(2!) + 3(3!) + ... + n(n!) = (n + 1)! - 1

You must show your working out either as a screenshot or by LaTeX.
Full solutions will be posted next week, along with a new question.

User Mroman
by
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1 Answer

3 votes
Let
n=1. Then the statement says


1(1!)=1

(1+1)!-1=2!-1=2-1=1

so it holds for the base case.

Assume it holds for
n=k, i.e. that


1(1!)+2(2!)+3(3!)+\cdots+k(k!)=(k+1)!-1

and use this to show it holds for
n=k+1. You have


1(1!)+2(2!)+\cdots+k(k!)+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!

=(k+1)!(1+k+1)-1

=(k+1)!(k+2)-1

=(k+2)!-1

which is what you needed to show, so the statement is true.
User Esko
by
8.8k points
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