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
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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.
(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.
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