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:

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

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