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I'm currently working on this problem which includes adding and multiplying (and probably factoring) factorials:

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

I need to make the left hand side equal to the right hand side without manipulating the right hand side. I've tried looking up methods on how to do this, but every example I've been able to find only showed cases where the denominator and numerator had factorials.

I know that $(k+1)!$ has to factor somehow, but I honestly don't know how to do it with factorials in this case.

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Factor the $(k+1)!$ from the LHS:

LHS=

$(k+1)!(1+k+1)-1$

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

=$(k+2)!-1$$\quad$ {because $(k+2)! = (k+1)!(k+2)$}

= RHS

QED

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  • $\begingroup$ The line that says (k+1)!(1+k+1)-1 had me confused for a second, but I think my understanding of it is right. I factored it in a less attractive way, but my -1 wound up within the parenthesis. I don't think that should cause problems when I write it like so: (k+1)!(1 + (k+1)(1) - 1) EDIT: My way actually does cause issues. How did you bring the -1 outside of the parenthesis? $\endgroup$ – Nick Sabia Apr 6 at 15:35
  • $\begingroup$ Just bring the -1 to the right and factor the first two terms. $\endgroup$ – aman Apr 6 at 15:38
  • $\begingroup$ OH I see, since we can't factor (k+1)! out of -1, we can leave it out of the parenthesis? $\endgroup$ – Nick Sabia Apr 6 at 15:39
  • $\begingroup$ (k+1)! -1+(k+1)(k+1)! = (k+1)! + (k+1)(k+1)! -1 $\endgroup$ – aman Apr 6 at 15:40
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    $\begingroup$ Then factor the (k+1)! from the first two terms $\endgroup$ – aman Apr 6 at 15:40

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