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(k+1)(k+1)!

Hi, as part of a proof through induction question, I have (k+1)(k+1)!, the answer for which is given as 1+(k+1)!, could anyone please explain why this is the case?

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Could you elaborate a bit more, because what you have described is super vague...
Reply 2
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bojbtij
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Original post by Star-girl
Could you elaborate a bit more, because what you have described is super vague...


Ok, sorry for the confusion: (k+1)(k+1)!= 1+(k+1)! but how is that the case
Original post by bojbtij
Ok, sorry for the confusion: (k+1)(k+1)!= 1+(k+1)! but how is that the case


It isn't.

Would you mind posting the whole question?
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bojbtij
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I would but I have no idea how to use sigma notation on here

Original post by Star-girl
It isn't.

Would you mind posting the whole question?
Original post by bojbtij
I would but I have no idea how to use sigma notation on here


http://www.thestudentroom.co.uk/wiki/LaTex
Reply 6
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bojbtij
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Prove through induction that ∑r=1nr∗r!=(n+1)!−1\displaystyle\sum_{r=1}^n r*r! =(n+1)!-1
(edited 8 years ago)
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Original post by Star-girl


Sorry forgot to quote
Original post by bojbtij
Prove through induction that ∑r=1nr∗r!=(n+1)!−1\displaystyle\sum_{r=1}^n r*r! =(n+1)!-1


If you add the (k+1)th term to the right hand side (subbing n for k) then can you go from there?
Original post by bojbtij
Prove through induction that ∑r=1nr∗r!=(n+1)!−1\displaystyle\sum_{r=1}^n r*r! =(n+1)!-1


OK. So what have you done so far? Work me through and we can maybe see where you might have gone wrong. :smile:
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Yes thank you both, a bit of an issue going from (k+1)!(k+2)-1 to (k+2)!-1 though

Original post by Wunderbarr
If you add the (k+1)th term to the right hand side (subbing n for k) then can you go from there?
Reply 11
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Original post by Star-girl
OK. So what have you done so far? Work me through and we can maybe see where you might have gone wrong. :smile:



All is well up until the last step i.e. (k+1)!(k+2)-1 to (k+2)!-1
Original post by bojbtij
Yes thank you both, a bit of an issue going from (k+1)!(k+2)-1 to (k+2)!-1 though


What is the definition of n!?
Reply 13
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bojbtij
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n*(n-1)*(n-2)...*3*2*1?

Original post by 16Characters....
What is the definition of n!?
(edited 8 years ago)
Original post by bojbtij
n+(n-1)+(n-2)+...+3+2+1?


Not quite, n!=n(n−1)(n−2)...1 n! = n(n-1)(n-2)...1 . So multiplication not addition.
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Original post by 16Characters....
What is the definition of n!?


Oh right, got it, thank you!
Reply 16
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Original post by 16Characters....
Not quite, n!=n(n−1)(n−2)...1 n! = n(n-1)(n-2)...1 . So multiplication not addition.


Yeah sorry, bit of a lapse
Reply 17
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Original post by bojbtij
All is well up until the last step i.e. (k+1)!(k+2)-1 to (k+2)!-1


(k+1)!=?(k+1)!=?
Original post by bojbtij
Yes thank you both, a bit of an issue going from (k+1)!(k+2)-1 to (k+2)!-1 though


You haven't done anything wrong - there is just a small thing you haven't realised. What is (k+2)(k+1)! ? Think about it.
Reply 19
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Original post by Star-girl
You haven't done anything wrong - there is just a small thing you haven't realised. What is (k+2)(k+1)! ? Think about it.


Original post by Andy98
(k+1)!=?(k+1)!=?



Yeah thanks it's just (k+2)(k+1)...3(2)(1), I think

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