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# Factorials

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Find the value of n that satisfies 2(n+1)!+6n!=4(n+1)!, where n! = n\cdot (n-1)\cdot (n-2) \cdots 2\cdot 1.

May 1, 2022

#1
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Use the latex button and enter your question properly.

May 1, 2022
#2
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We have the equation: $$2(n+1)!+6n!=4(n+1)!$$

We can subtract $$2(n+1)!$$ from both sides, which yields: $$6n!=2(n+1)!$$

Note that $$(n+1)! = n! \times (n+1)$$

Substituting this gives us: $$6n!=2(n+1)n!$$

Now, we just have to divide by $$n!$$, then solve for n.

Can you take it from here?

May 1, 2022