# Solve the equation (n+1)! + n! = 35(n-1)!

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

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

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

=> n + 1 + 1 = 35/n

=> n + 2 = 35/n

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

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

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

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

So n can be 5 and -7

But as the factorial of a negative number is not defined we consider only n+ 5

**Therefore n = 5.**

To solve the equation (n+1)! + n! = 35(n-1)!

We know that n! = 1*2*3*4*...*n.

Therefore (n+1)! = (n+1)*n!. And n! = n*(n-1)!.

So (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 =0

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

Setting each factor to zero, we get:

So n = 5)*n!.

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.**