# SolutionsFind the solutions of the equation: (n+1)!+n!=35(n-1)!

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### 2 Answers

We have to solve (n+1)!+n!=35(n-1)!

n! = 1*2*3...*n

(n + 1)! = (n + 1)(n)(n - 1)!

n! = n(n - 1)!

(n+1)!+n!=35(n-1)!

=> (n + 1)(n)(n - 1)! + n(n - 1)! = 35 (n - 1)!

=> (n + 1)*n + n = 35

=> n^2 + n + n = 35

=> n^2 + 2n = 35

=> n^2 + 2n - 35 = 0

=> n^2 + 7n - 5n - 35 = 0

=> n(n + 7) - 5(n + 7) = 0

=> (n - 5)(n + 7) = 0

n = 5 or n = -7

But the factorial is defined for only positive values so we eliminate x = -7

**The required solution is n = 5**

We'll write (n+1)! and n!, with respect to (n-1)!.

(n+1)! = (n-1)!*n*(n+1)

n! = (n-1)!*n

We'll re-write the equation:

(n-1)!*n*(n+1) + (n-1)!*n = 35(n-1)!

We'll factorize by (n-1)!

n(n-1)![(n+1) + 1] = 35(n-1)!

We'll divide by (n-1)!:

n(n+2) = 35

We'll remove the brackets:

n^2 + 2n - 35 = 0

We'll apply the quadratic formula:

n1 = [-2+sqrt(4 + 140)]/2

n1 = (-2 + 12)/2

n1 = 5

n2 = (-2-12)/2

n2 = -7

**Since n has to be a natural number, we'll reject the second solution and we'll keep n = 5.**