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Find the value of n that satisfies 2(n+1)!+6n!=3(n+1)!, where n! = n * (n-1) * (n-2)...2*1.

 

The answer is not 0.

 Jul 27, 2020
 #1
avatar+21953 
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Knowing that  (n + 1)!  =  (n + 1)·n!:

 

2(n + 1)!  +  6n!  =  3(n + 1)!

 

2·(n + 1)·n!  +  6·n!  =  3·(n + 1)·n!

 

Divide both sides by  n!:

 

2(n + 1)  +  6  =  3(n + 1)

    2n + 2 + 6  =  3n + 3

          2n + 8  =  3n + 3

                  8  =  n + 3

                  5  =  n

 Jul 27, 2020

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