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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
avatar+1366 
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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

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