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Which of the following is equal to n!*(n+1)!?
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12 Oct 2021, 10:03
12 Oct 2021, 10:03
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81% (00:53) correct
18% (02:36) wrong based on 16 sessions
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Which of the following is equal to n!*(n+1)!?
A. \((n!)^2\)
B. \((n+1)!^2\)
C. \(n!^3(n+1)!^2\)
D. \(n!^3(n+1)\)
E. \(n!^2(n+1)\)
A. \((n!)^2\)
B. \((n+1)!^2\)
C. \(n!^3(n+1)!^2\)
D. \(n!^3(n+1)\)
E. \(n!^2(n+1)\)
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Re: Which of the following is equal to n!*(n+1)!?
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12 Oct 2021, 10:07
12 Oct 2021, 10:07
Carcass wrote:
Which of the following is equal to n!*(n+1)!?
A. \((n!)^2\)
B. \((n+1)!^2\)
C. \(n!^3(n+1)!^2\)
D. \(n!^3(n+1)\)
E. \(n!^2(n+1)\)
A. \((n!)^2\)
B. \((n+1)!^2\)
C. \(n!^3(n+1)!^2\)
D. \(n!^3(n+1)\)
E. \(n!^2(n+1)\)
Notice that (n+1)(n!) = (n+1)!
Why is this?
Well if we expand n! to be (n)(n-1)(n-2)....(3)(2)(1) we can see why.
We get: (n+1)(n!) = (n+1)(n)(n-1)(n-2)....(3)(2)(1)
= (n+1)!
So, (n!)(n+1)! = (n!)(n+1)(n!)
= (n!)² (n+1)
Answer: E
General Discussion
Re: Which of the following is equal to n!*(n+1)!?
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23 Feb 2023, 01:16
23 Feb 2023, 01:16
Hi Brent GreenlightTestPrep
So at this stage (n!) and (n)(n-1)(n-2)....(3)(2)(1) cancel out so we are left with (n+1)!
We get: (n+1)(n!) = (n+1)(n)(n-1)(n-2)....(3)(2)(1)
= (n+1)!
But then not sure how/why is additional (n!) being multiplied here? Could you help clarify? Thanks
So, (n!)(n+1)! = (n!)(n+1)(n!)
= (n!)² (n+1)
So at this stage (n!) and (n)(n-1)(n-2)....(3)(2)(1) cancel out so we are left with (n+1)!
We get: (n+1)(n!) = (n+1)(n)(n-1)(n-2)....(3)(2)(1)
= (n+1)!
But then not sure how/why is additional (n!) being multiplied here? Could you help clarify? Thanks
So, (n!)(n+1)! = (n!)(n+1)(n!)
= (n!)² (n+1)
