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

Предыдущий
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вопрос

Let's rewrite the left side of the equation:

(n-1)! + n! + (n+1)!
= n! + (n-1)! (1+n+n(n+1))
= n! + (n-1)! (n+1) (n+1)
= n! + n! (n-1) (n+1)
= n! (1 + (n-1)(n+1))
= n! (1 + (n^2 - 1))
= n! (n^2)
= n! * n^2

Then, the left side of the equation becomes:

n! n^2 = (n+1)^2 (n-1)!
= (n+1)^2(n-1)!.

Therefore, the equation (n-1)! + n! + (n+1)! = (n+1)^2(n-1)! is proven to be true.