Show that #(n+2)!+(n+1)!+n!##=n!(n+2)^2# ?

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Answer 1 · 2017-10-10T23:10:17

See explanation

Start with the expression on the left:

#(n+2)!##+(n+1)!+n!#

Simplify using factorials:

#(n+2)(n+1)n!+(n+1)n!+n!#

Factor out the #n!#:

#n!((n+2)(n+1)+(n+1)+1)#

Simplify inside the parentheses:

#n!(n^2+3n+2+n+2)#

#n!(n^2+4n+4)#

Factor:

#n!(n+2)^2#

So:

#(n+2)!##+(n+1)!+n!##=##n!(n+2)^2#