Answered

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

Sagot :

MrRoyal

Answer:

[tex]n = 5[/tex]

Step-by-step explanation:

Given

[tex]2(n+1)! + 6n! = 3(n+1)![/tex]

Required

Find n

Simplify (n + 1)!

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

Factorize

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

Divide both sides by n!

[tex]2(n+1) + 6 = 3(n+1)[/tex]

Open brackets

[tex]2n + 2 + 6 = 3n + 3[/tex]

[tex]2n + 8 = 3n + 3[/tex]

Collect like terms

[tex]2n - 3n = 3 -8[/tex]

[tex]-n =-5[/tex]

Multiply both sides by -1

[tex]n = 5[/tex]

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