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Mathematical Induction Problem - Please Help With This Question (1 Viewer)

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Hi guys just need some help with an induction problem. Got a sheet we gotta have done by friday with a heap of questions on it. Done them all except this one now no matter what i do i cant figure out how to do it. I've even the got worked solution but it still doesnt help me. I can do it all except for one line in the working i cant understand. Any help would be greatly appreciated. The problem is in the attached text file with the solution underneath and an explanation of where im having trouble. Any help you could give me is greatly appreciated. You can either reply here with help or email at:

stephen_bradley@iinet.com.au

Thanks again cya
 
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Oops sorry heres the attachment

Hey sorry forgot to add the attachment with the problem here it is now
 
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By the way if your interested the question is in the 1999 3 unit paper - question 5a
 

Slidey

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Well, when you multiply the top by (k+1) you also have to divide the bottom by (k+1) to maintain equality.

EDIT: BTW, if this happen in the future, my best advice is to leave it, go for a jog, do some other homework, sleep on it, whatever and come back to it when you're in a different mindset - you'll get it straight away. Some would say your brain subconsciously thinks about it while you're doing the other activities. I'd tend to agree.
 
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Hey i still dont get it, where is it multiplied by k+1 on the top i thought u would just take:
Sum k*Term(k+1)
And this would be (k+1)(k+2)(k+3)…(2k-1)2k(2k+1)(2k+2) what am i doing wrong? if possible can anyone give me a worked solution to the answer explaining in detail each step or even just a worked solution by doing it another way altogethter
 
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yeah i bothered i really need the help its part of a formal assesment take home assignment thingy so i need to figure out how to do it coz she knows theyre in the book and i need to write it out as if i was doin it so i need to understand it
 

Slidey

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Code:
Now LHS   = (k+2)(k+3)…(2k+1)(2k+2)
		
          = [u](k+1)(k+2)(k+3)…(2k-1)2k(2k+1)(2k+2)[/u]
		    	k+1
First step: you start with (k+2), not (k+1)
Second step: you start with (k+1) instead of (k+2) - you've multiplied by (k+1), but to balance this, you divide by (k+2).
 

dawso

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yeah, u are thinking along the right lines but this solution has skipped steps (which are easy to skip cause its easy)

(n+1)(n+2)...(2n-1)2n

n=k+1

((k+1)+1)((k+1)+2)...(2(k+1)-1)2(k+1)
= (k+2)(k+3)...(2k+1)(2k+2)
then they divide/multiply by k+1......

i think this is the bit u cant get, if not, i suppose u will need a full solution, or better still, just drop 3unit....
 

Slidey

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dawso said:
i think this is the bit u cant get, if not, i suppose u will need a full solution, or better still, just drop 3unit....
Please do not suggest inconsiderate things like that - at all.

One, you don't know anything about this person's ability or interest in maths and two, simply because they don't get a solution, or if they make a silly mistake, this is not cause for them to drop 3u.
 

dawso

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im sorry mother, just remember boys, i did attempt to help him first, anyway crippled kunfu, do u get this now?
 
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Hey thanks guys i get it now, once i tried o do it when it wasnt after 10 at night it wasnt 2 hard lol.
 

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