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Mathematics Inclined Week #1: Difficult Category

Problem:
Prove, by mathematical induction, that:

[tex]1(1!) + 2(2!) + 3(3!) + ... + n(n!) = (n + 1)! - 1[/tex]

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.

Respuesta :

LammettHash
Let [tex]n=1[/tex]. Then the statement says

[tex]1(1!)=1[/tex]
[tex](1+1)!-1=2!-1=2-1=1[/tex]

so it holds for the base case.

Assume it holds for [tex]n=k[/tex], i.e. that

[tex]1(1!)+2(2!)+3(3!)+\cdots+k(k!)=(k+1)!-1[/tex]

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

[tex]1(1!)+2(2!)+\cdots+k(k!)+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)![/tex]
[tex]=(k+1)!(1+k+1)-1[/tex]
[tex]=(k+1)!(k+2)-1[/tex]
[tex]=(k+2)!-1[/tex]

which is what you needed to show, so the statement is true.

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