Solve the expression ( n +1 )!/( n-1)! = 30
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(n+1)!/(n-1)!=30
First let us simplify:
(n+1)(n)(n-1)!/(n-1)1=30
Now reduce similar:
==? (n+1)n=30
==> n^2+n-30=0
==> (n+6)(n-5)=0
==> n={-6,5}
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The equation (n+1)!/(n-1)! = 30 has to be solved.
The factorial of a number n, n! is equal to the product of all positive integers from 1 till n, n included.
(n+1)! = (n+1)*n*(n-1).... 1
(n-1)! = (n-1)*(n - 2)* ... 1
This gives:
((n+1)*n*(n-1).... 1)/((n-1)*(n - 2)* ... 1) = 30
(n+1)*n = 30
n^2 + n - 30 = 0
n^2 + 6n - 5n - 30 = 0
n(n + 6) - 5(n+6) = 0
(n - 5)(n + 6) = 0
n = 5 and n = -6
But the factorial for negative numbers is not defined, eliminate n = -6
The required solution of the equation is n = 5.
To solve for (n+1)!/(n-1)! = 30.
Solution:
(n+1)! = (n+1)(n)(n-1)(n-2)(n-3)....(3)(2)(1)
Threrefore, (n+1)! = (n+1)(n)*(n-1)!. By dividing both sides , we get :
(n+1)!/(n-1)! = (n+1)(n). Using this on the LHS of the given equation, we get:
(n+1)n = 30. Or
6*5 = 30. So, n = 5 .
(n+1)!/(n-1)! =
(n+1)!= 1x2x3x...(n-1)x(n)x(n+1)
(n-1)!= 1x2x3x...(n-1)
(n+1)!/(n-1)!= 1x2x3x...(n-1)x(n)x(n+1)/1x2x3x...(n-1)
We can simplify the same factors from numerator and denominator: 1x2x3x...(n-1)
Simplifying, it will remain at numerator (n)x(n+1)
(n)x(n+1)= 30
n^2 +n= 30
n^2 +n -30= 0
n1={-1+ [1-4x(-30)]^1/2}/2= (-1 + 121^1/2)/2= (-1 +11)/2
n1=5
n2 = {-1- [1-4x(-30)]^1/2}/2= (-1 - 121^1/2)/2= (-1 -11)/2
n2=-6
Both solution are natural numbers
n= {-6;5}
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