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

 Feb 13, 2021
 #1
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2 ( n+ 1) !   +  6n!   =  3 ( n + 1)!

 

Note  that  (  n + 1)! =    (n + 1) * n!

 

So

 

2 (n + 1) * n!    +  6n!  =  3 (n + 1) * n!            divide out  n!

 

2 ( n + 1)  +   6  =  3 ( n + 1)        simplify

 

2n + 2  +  6 =  3n  +  3

 

8 =  n +  3

 

8  - 3   = n  =   5

 

 

cool cool cool

 Feb 13, 2021

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