Hi there,

I understand that the numerator becomes (n!)2 -1, and the denominator becomes n! +1, but why does the 2^2 disappear? Do we not still have two on the numerator since there is only 1 2 in the denominator?

Thanks for your help in advance.

C

Hint: first show that the problem can be simplified to

\frac{(n!)^2 - 1}{n! + 1}

Then think about how you can use the a^2 - b^2 = (a + b)(a - b) identity to finish off the problem.

I am not able to simplify the equation to what you listed above. So far I can see 1^2-1 = (1-1)(1+1) in the numerator and in the denominator you have (1+1).

Therefore, the (1+1) will be cancelled and you are left with (1-1) in the numerator which equals 0. So 2^2 * 0 = 0.

With my approach I still end up with a 2 in the denominator and the -1 in the numerator has disappeared.

Where did I go wrong?

\dfrac{(n!+1)(n!-1)}{(n!+1)} = 119

\dfrac{\cancel{ (n!+1)} (n!-1)} {\cancel{(n!+1)}}=119

n!-1=119

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