SOLUTION: im not sure what this falls under. it has been years since ive done alg. :
The question ask: For any positive integer n, (((n+1)!)/(n!))-n = ?
answers: (a) 0 (b) 1 (c)
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The question ask: For any positive integer n, (((n+1)!)/(n!))-n = ?
answers: (a) 0 (b) 1 (c)
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Question 181096: im not sure what this falls under. it has been years since ive done alg. :
The question ask: For any positive integer n, (((n+1)!)/(n!))-n = ?
answers: (a) 0 (b) 1 (c) n (d) n+1 (e) n!
I really dont remember what the ! stands for so i dont know how to go about looking for help. Found 2 solutions by Alan3354, Mathtut:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! im not sure what this falls under. it has been years since ive done alg. :
The question ask: For any positive integer n, (((n+1)!)/(n!))-n = ?
answers: (a) 0 (b) 1 (c) n (d) n+1 (e) n!
I really dont remember what the ! stands for so i dont know how to go about looking for help.
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! means factorial.
It's n*(n-1)*(n-2)*(n-3) ... 1
For example, 6! = 6*5*4*3*2*1 = 720
Most calculators have this function.
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So, n!/(n-1)! = n.
Then subtracting n gives 0.
It's a, zero.
You can put this solution on YOUR website! ! is factorial
which means 4! is 4*3*2*1
:
so in this problem you would be left with
:
n+1-n which equals 1
:
so the answer is b
:
((n+1)*n*(n-1)*(n-2)...)/(n*(n-1)(n-2)....))-n
:
notice how everything from the n onward cancels so we are left with
:
(n+1)-n
(