Question 856666: Show that (n+1)!-3(n!)+(n-1)!=(n-1)!(n-1)^2
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Keep in mind that n! = n*(n-1)*(n-2)...(3)*(2)*(1)
You can shorten this to say n! = n(n-1)!
Similarly, (n+1)! = (n+1)*n! which leads to (n+1)! = (n+1)*n*(n-1)!
We'll use these equations to make substitutions below
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(n+1)!-3(n!)+(n-1)!=(n-1)!(n-1)^2
(n+1)*(n)*(n-1)!-3(n)*(n-1)!+(n-1)!=(n-1)!(n-1)^2
(n-1)! [ (n+1)*(n)-3(n)+1 ] = (n-1)!(n-1)^2
(n-1)! [ n^2+n-3n+1 ] = (n-1)!(n-1)^2
(n-1)! [ n^2-2n+1 ] = (n-1)!(n-1)^2
(n-1)! (n-1)^2 = (n-1)!(n-1)^2
So the identity is confirmed.
Notice how I only changed the left side and didn't alter the right side.
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