# Solve the equation n! + (n+1)! = (n+2)!.

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n! + (n+1)! = (n+2)!

Let us rewrite:

n!+ (n+1)n! = (n+2)(n+1)n!

Factorize n!

n!(1+n+1) = (n+2)(n+1)n!

Divide by n!

==> n+2 = (n+2)(n+1)

No divide by (n+2)

==> 1= n+1

==> n = 0

to verify substitute with n=0

==> 0! + 1! = 2!

==> 1+1 = 2

==> 2=2

(n+2)!= 1*2*3*...(n-1)*(n)*(n+1)*(n+2) = n!*(n+1)*(n+2)

(n+1)!= 1*2*3*...(n-1)*(n)*(n+1) = n!*(n+1)

n!= 1*2*3*..*(n-1)*n

n! + (n+1)! = (n+2)!

We'll re-write the equation:

n! + n!*(n+1) = n!*(n+1)*(n+2)

We'll factorize, to the left side:

n!*(1+n+1) = n!*(n+1)*(n+2)

n!*(n+2) = n!*(n+1)*(n+2)

We can divide by the same factors, from both sides.

1 = n+1

We'll add the value -1, both sides:

1-1 = n+1-1

**n=0, is the solution of the equation.**

The equation n! + (n+1)! = (n+2)! has to be solved for n.

The factorial of a number n! is defined by the product n*(n-1)*(n-2)...(1)

This gives two relations very useful in solving the given equation.

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

(n+2)! = (n+2)(n+1)*(n)(n-1)...(1) = (n+2)*(n+1)*n!

Substitute these in the given equation. He have:

n! + (n+1)*n! = (n+2)*(n+1)*n!

Cancel n!

1 + (n+1) = (n+2)(n+1)

n+2 = (n+1)(n+2)

n+1 = 1

n = 0

This gives the solution n = 0 but remember that the factorial is defined only for positive numbers and 0! is given a value of 1.

To solve n!+(n+1)! = n+2)!.............(1)

Solution:

We know that if n is a positive integer , then n! = 1*2*3*4....(n-2)(n-1)n.

So, (n+2)! = (n+2)(n+1)*n! and

(n+1)! = (n+1)n!. Substituting in (1), we get:

n! +(n+1)n! = (n+2(n+1)n!, So we divide by n! both sides:

1+(n+1) = (n+2)(n+1) = n^2+3n+2. Or

n+2 = n^2+3n+2.Or

0 = n^2+2n

n(n+2) = 0. Or n = 0.Or n+2 = 0. Or n-2. This contradicts n is a positive integer. Hence n = 0 is the only solution.