Prove $(n-2)!+(n-1)!+n!=(n-2)!n^2$ for $n\ge 2$.

Question

Sorry I'm not sure how do the symbols such as the same as or powers. This is probably a very easy question but I can only think of proving it by proof of exhaustion. However as $n$ is basically infinite above $2$ I don't think this is right.

Thanks for any help!

Answer 1

You can take a factor from $(n-2)!$ likes the following: $$(n-2)! + (n-1)! + n! = (n-2)!(1 + (n-1) + (n-1)\times n) = (n-2)!(1 + (n-1)\times(n+1)) = (n-2)!(1 + n^2 - 1) = (n-2)! n^2$$

Answer 2

You know that: $$n!=n\cdot(n-1)\cdot (n-2)\cdot(n-3)\ldots$$ So therefore: $$n!=n\cdot (n-1)\cdot(n-2)!=(n^2-n)\cdot(n-2)!$$ and: $$(n-1)!=(n-1)\cdot(n-2)!$$ Putting it all together: $$n!+(n-1)!+(n-2)!=(n^2-n)\cdot(n-2)!+(n-1)\cdot(n-2)!+(n-2)!=(n^2-n+n-1+1)(n-2)!=n^2(n-2)!$$

Answer 3

we have $$n!=n(n−1)!=n(n−1)(n−2)! = n^2(n−2)! -n(n−2)!$$ then, $$ (n−2)!+(n−1)!+n! = (n−2)!+(n−1)(n−2)!+n^2(n−2)! -n(n−2)! = n^2(n−2)! +\underbrace{(n−2)!-n(n−2)!+(n−1)(n−2)!}_{0} =n^2(n−2)! $$