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FP1 Induction X1

I really can't get my head around the harder kind of these summation problems. Especially this factorial one.

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I don't think it's likely something like this will appear in the 2016 FP1 paper but I have to make sure I would know how to deal with this just in case.

Thanks




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Original post by chemistryboi
I really can't get my head around the harder kind of these summation problems. Especially this factorial one.

Attachment not found


I don't think it's likely something like this will appear in the 2016 FP1 paper but I have to make sure I would know how to deal with this just in case.

Thanks




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What have you tried so far? :h:
@seanfm

I have currently shown the summation formula is true for n=1

(1)1=(1+1)!-1
1=(2x1)-1=1

I've assumed true for n=k but not much further than that apart from substituting n=k+1


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Reply 3
the attachment does not work
Original post by TeeEm
the attachment does not work


Sigma has n on the top and r=1 at bottom

Sigma r(r!) = (n+1)!-1


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Original post by chemistryboi
@seanfm

I have currently shown the summation formula is true for n=1

(1)1=(1+1)!-1
1=(2x1)-1=1

I've assumed true for n=k but not much further than that apart from substituting n=k+1


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What does your substitution look like for n=k+1, i.e what have you done?
Original post by SeanFM
What does your substitution look like for n=k+1, i.e what have you done?


ImageUploadedByStudent Room1454524860.368800.jpg

It feels painfully obvious what I have to do next but I can't piece it together


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Original post by TeeEm
the attachment does not work


I have added a new picture, using the student room app atm and it's very buggy.


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Original post by chemistryboi
ImageUploadedByStudent Room1454524860.368800.jpg

It feels painfully obvious what I have to do next but I can't piece it together


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If you move the -1 to the very right hand side of the equation at the bottom, can you see anything that you can do to the first two terms?
Reply 9
Original post by chemistryboi
I have added a new picture, using the student room app atm and it's very buggy.


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If everything else fails take a look at this link
http://madasmaths.com/archive/maths_booklets/further_topics/various/proof_by_induction.pdf
Original post by SeanFM
If you move the -1 to the very right hand side of the equation at the bottom, can you see anything that you can do to the first two terms?


i tried that and factorised to get

(k+1)![1+(k+1)]-1 = (k+1)!(k+2)-1
Original post by xyz9856
i tried that and factorised to get

(k+1)![1+(k+1)]-1 = (k+1)!(k+2)-1


Awesome! You are almost there, there are just... two steps for the proving it for n=k+1 and then the usual 'by mathematical induction'.. stuff.


Wow what an excellent set of resources, I may have to try out some of those papers on the site. Thanks for sharing that with me
Original post by SeanFM
Awesome! You are almost there, there are just... two steps for the proving it for n=k+1 and then the usual 'by mathematical induction'.. stuff.


I can't see it, its so frustrating, any hints?
Reply 14
Original post by xyz9856
Wow what an excellent set of resources, I may have to try out some of those papers on the site. Thanks for sharing that with me


my pleasure
Original post by xyz9856
I can't see it, its so frustrating, any hints?


Oh, sorry, you are one step away :tongue: as I did not see the bit after = which was.

[(k+1)!(k+2)] -1.

Remember that you are looking for it to be of the form (n+1)! - 1, where n = k+1..

What can (k+1)! * (k+2) be turned into? (Hint: it's another factorial).
Original post by SeanFM
Oh, sorry, you are one step away :tongue: as I did not see the bit after = which was.

[(k+1)!(k+2)] -1.

Remember that you are looking for it to be of the form (n+1)! - 1, where n = k+1..

What can (k+1)! * (k+2) be turned into? (Hint: it's another factorial).


im guessing it gets turned into (k+2)! but I don't really understand why
Original post by xyz9856
im guessing it gets turned into (k+2)! but I don't really understand why


Precisely.

The definition of (k+2)! for some k is (k+2)*(k+1)*(k)*(k-1)....2*1. and the defintion of (k+1)! is (k+1)*(k)*(k-1)...2*1

so (k+2)*(k+1)! is..
Original post by SeanFM
Precisely.

The definition of (k+2)! for some k is (k+2)*(k+1)*(k)*(k-1)....2*1. and the defintion of (k+1)! is (k+1)*(k)*(k-1)...2*1

so (k+2)*(k+1)! is..


that took a lot of staring to understand, but it is blindingly obvious now.

(k+2)*(k+1)(k)(k-1) is clearly (k+2)!

sorry for the late reply and thank you very much for your help
Original post by xyz9856
that took a lot of staring to understand, but it is blindingly obvious now.

(k+2)*(k+1)(k)(k-1) is clearly (k+2)!

sorry for the late reply and thank you very much for your help


No worries.

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