Question

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

Answer 1

`n = 5`

Explanation:

Given

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

Required

Find n

Simplify (n + 1)!

`2(n+1)*n! + 6n! = 3(n+1)*n!`

Factorize

`n![2(n+1) + 6] = 3(n+1)*n!`

Divide both sides by n!

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

Open brackets

`2n + 2 + 6 = 3n + 3`

`2n + 8 = 3n + 3`

Collect like terms

`2n - 3n = 3 -8`

`-n =-5`

Multiply both sides by -1

`n = 5`