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

Answer:

$n = 5$

Step-by-step 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

$n = 5$