That_Guy_Omzz
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Why is (k + 1)[(k + 1) !] = [1 + k + 1] ?
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Plücker
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(Original post by That_Guy_Omzz)
Why is (k + 1)[(k + 1) !] = [1 + k + 1] ?
It isn't.
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0le
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Maybe the question is asking you to prove or disprove the claim. You can disprove the claim in this case with a simple counterexample. Assume k is in the set of natural numbers, and try some values of k.
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Kallisto
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(Original post by That_Guy_Omzz)
Why is (k + 1)[(k + 1) !] = [1 + k + 1] ?
It is no such thing.

Examples: (0+1)! = 1! = 1 for k = 0; (1 + 1)! = 2! = 1*2 = 2 for k = 1; (2 + 1)! = 3! = 1*2*3 = 6 for k = 2

What do you think is (k + 1)! when you write it as a multiplication chain?
Last edited by Kallisto; 1 month ago
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That_Guy_Omzz
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Take a look
Attached files
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15Characters...
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(Original post by That_Guy_Omzz)
Take a look
That attachment says that:

(k+1)! + (k+1)[(k+1)!] - 1 = (k+1)![1 + k + 1] - 1

where we factorised the (k+1)! out of the first two terms.


Note that this is not what you put in your original post.
Last edited by 15Characters...; 1 month ago
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davros
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(Original post by That_Guy_Omzz)
Take a look
This is just grouping terms with a common factor.

You have a (k+1)! multiplied by (k+1) and another (k+1)! on its own (i.e multiplied by 1)

So in total you have lk + 1 + 1 lots of (k+1)!
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0le
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Damn, this is part of A-Level syllabus? I suddenly feel very dumb and old
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RogerOxon
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(Original post by That_Guy_Omzz)
Take a look
It says:
(k+1)!+(k+1)!(k+1)=(k+1)!(k+2)=(  k+2)!
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That_Guy_Omzz
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(Original post by 0le)
Damn, this is part of A-Level syllabus? I suddenly feel very dumb and old
Further Maths, don't worry about it lol
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