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OpenStudy (a_clan):
the expression (n+1)!/(n-1)! should be part of an equality
OpenStudy (a_clan):
(n+1)!/(n-1)! = RHS
OpenStudy (anonymous):
RHS?
OpenStudy (a_clan):
right hand side
(LHS) x = y (RHS)
OpenStudy (anonymous):
try n^2+n
put values
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OpenStudy (anonymous):
(n+1)!/(n-1)!=20
n!=?
OpenStudy (anonymous):
any1?
OpenStudy (anonymous):
like I said, try n^2+n
OpenStudy (a_clan):
The trivial method is to find 'n' and then solve n! , as panlac01 said.
Method 1:
(n+1)!/(n-1)!=20
(n+1)n. (n-1)! / (n-1)! = 20
(n+1)n = 20
or n^2 + n = 20
OpenStudy (anonymous):
solve for N for the answer?
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OpenStudy (a_clan):
OR
Method 2:
(n+1)!/(n-1)! = (n+1) n! / (n-1)! .....................eqn (A)
||
20 = 5*4
= 5 * 4 *(3!) / (3!) [Multiply and divide by 3!]
= 5 * 4! / 3!
[Now this is the same form as the RHS of eqn (A)]
You can compare n! = 4! =24 (SOLUTION}
OpenStudy (anonymous):
thank u. the "!" confuesd me. thanx for the help, bro!