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Mathematical Induction

  1. Jul 10, 2008 #1
    Hi this is my first post so here goes.....

    Basically I'm studying maths and in a section called proof and resoning they have introduced mathematical induction. I have tried to follow the examples but I still can't make head nor tail of it really. It makes absolutely no sense to me at all?

    The question asked is

    Prove SUM(r=1 to n) 1/r(r + 1)(r + 2) is 1/4 - 1/2(n + 1)(n + 2)


    Ok where/how do I start? This bears absolutely no resemblance to any of the worked examples at all, the examples are summing squares and cubes but this is a fraction?
    Can you see my problem? If I'm absolutely honest I have been on with it for a few days now (yes I said days!) and am about to forget the whole induction thing completely...


    Please if anyone can help me have a light bulb moment I will be very grateful!!!

    Many thanks


    James
     
  2. jcsd
  3. Jul 10, 2008 #2

    HallsofIvy

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    A nice thing about induction is that you always start the same way: show that the statement is true for n= 1. For n= 1, the left hand side is just the one term 1/(1(1+1)(1+2))= 1/6 while the right hand side is 1/4- 1/(2(1+1)(1+2))= 1/4- 1/12= 3/12- 1/12= 2/12= 1/6. Yes, the statement is true when n= 1.

    And, the second part is basically always the same: Assume the statement is true for n=k and prove that it must be true for n= k+1.

    What you WANT to show, then is the statement with n= k+1:
    [tex]\sum_{r=1}^{k+1} \frac{1}{r(r+1)(r+2)}= \frac{1}{4}- \frac{1}{2(n+1)(n+2)}[/tex]

    Okay, the left side is just the previous sum with one new term:
    [tex]\sum_{r=1}^{k}\frac{1}{r(r+1)(r+2)}+ \frac{1}{(k+1)(k+1+1)(k+2+1)}[/tex]

    Assuming that the formula is correct for n= k, that is the same as
    [tex]\frac{1}{2(k+1)(k+2)}+ \frac{1}{(k+1)(k+2)(k+3)}[/tex]

    Now, do the algebra. Get common denominators and add the fractions, then show that what you get is the same as I said above.
     
  4. Jul 10, 2008 #3
    Thanks for the reply! Well to be honest what you have said has meant more to me than the past few days have!
    When I get home I will try and work through it as you have said - and report back...


    Thanks again.

    James
     
  5. Jul 10, 2008 #4
    I've tried but failed again. I can't get

    1/2k(k + 1)(k + 2) + 1/(k + 1)(K + 2)(k + 3)

    to be

    1/2(n + 1)(n + 2)
     
  6. Jul 10, 2008 #5
    No it should be 1/(2*(n + 2)(n+3))...
     
  7. Jul 10, 2008 #6
    As you can see I'm finding this very difficult to understand.
    Can you tell me in 'idiot speak'? So I can try and get a grip of it...
     
  8. Jul 10, 2008 #7
    Which part are you not understanding. Do you understand the concepts of MI? Are you having more trouble manipulating the equation with the Algebra?
     
  9. Jul 10, 2008 #8

    HallsofIvy

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    That's 1/4- 1/(2k(k+1)(K+2)+ 1/[(k+1)(k+2)(k+3)]

    The "1/4" you can just leave but that "-" is important!
     
  10. Jul 11, 2008 #9
    To be honest I am having trouble undrstanding MI from top to bottom - the whole 9 yards basically!
    If its not too much hassle I would like a 'walk through' of it if at all possible...

    Thanks for your help so far

    James
     
  11. Jul 11, 2008 #10

    HallsofIvy

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    Are you capable of basic algebra?

    What is 1/4- 1/(2k(k+1)(K+2)+ 1/[(k+1)(k+2)(k+3)]?

    Watching me do for you wouldn't help you any more than my watching you do 30 pushups would help me!
     
  12. Jul 11, 2008 #11
    lol yeah I can do basic algebra! Please forgive me, I'm studying through the OU and this has just been introduced. I have previously done Further Pure. Personally I don't think that the 'concept' of what we are trying to do/achieve has not been explained very well. Bear in mind this 'technique' is new to me and I've had a mental block on this for a few days now. Once the light bulb comes on and I understand what is wanted and what to do and how to apply it to various 'situations' all will be well.

    Thanks

    James
     
  13. Jul 11, 2008 #12

    HallsofIvy

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    The best way to get ouot of that "mental block" is to do the algebra that I have suggested twice now: what is 1/4- 1/(2k(k+1)(K+2)+ 1/[(k+1)(k+2)(k+3)]?
     
  14. Jul 11, 2008 #13
    What am I supposed to end up with after doing the algebra?

    1/4 - 1/2(k + 1)(k +2) ....?
     
  15. Jul 11, 2008 #14
    Thanks for all the help so far - it is appreciated! I don't want to jump the gun but I think the penny may have dropped....
    Can someone please check my working?

    We have

    1/4 - 1/2k(k + 1)(k + 2) + 1/)k + 1)(k + 2)(k + 3)

    1/4 - 1/4*4/2k(k + 1)(k + 2) + 1/4*4/(k + 1)(k + 2)(k + 3)

    1/4{ 1/1 - 4/2k(k + 1)(k + 2) + 4/(k + 1)(k + 2)(k +3) }

    1/4{ (2k(k + 1)(k + 2)(k + 3) -4(k + 3) + 4(2k))/2k(k + 1)(K + 2)(k + 3)}

    1/4{ 2k^4 + 12k^3 + 22k^2 + 8k -12/2k(k + 1)(k + 2)(k + 3) }

    now divide the numerator by (k + 3) to arrive at

    1/4{ (k + 3)( 2k^3 + 6k^2 + 4k -4)/2k(k + 1)(k + 2)(k + 3) }

    Cancel the (k +3) top and bottom

    Take k out of the top

    1/4{ k(2k^2 + 6k + 4) -4 / 2k(k + 1)(k + 2) }

    1/4{ k(2k^2 + 6k + 4)/2k(k + 1)(k + 2) - 4/2k(k + 1)(k + 2) }

    1/4 { 2K^2 + 6k + 4/ 2(K + 1)(k + 2) - 2/k(k + 1)(k + 2) }

    1/4 { 1 - 2 / k(k + 1)(k +2) }

    1/4 - 1/ 2k(k + 1)(k +2 )

    Finally end up with what was asked?

    Many Thanks

    James
     
  16. Jul 11, 2008 #15

    HallsofIvy

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    The second fraction does not have a "k" in the denominator, it is just -1/[2(k+1)(k+2)]. That's my fault, I copied it wrong myself.

    I don't know why you have incorporated the "1/4" in that- it just "goes along for the ride" and, as I said before, you can just leave it.
    To combine -1/[2(k+1)(k+2)+ 1/(k+1)(k+2)(k+3) you need a common denominator of 2(k+1)(k+2)(k+3). That means that you need to multiply numerator and denominator of the first fraction by k+3 and of the second fraction by 2:
    -(k+3)/[2(k+1)(k+2)(k+3)]+ 2/[2(k+1)(k+2)(k+3)]= (-k- 3+ 2)/[2(k+1)(k+2)(k+3)]= (-k-1)/[2(k+1)(k+2)(k+3)]= -(k+1)/[2(k+1)(k+2)(k+3)]= -1/[2(k+2)(k+3)] so, putting the 1/4 back in
    that is 1/4- 1/[2(k+2)(k+3)].

    Can you see that that is the same as the orginal formula with n replaced by k+1?

     
  17. Jul 12, 2008 #16
    Thanks to all for your help - I've finally managed to understand what is being asked.
    Now that I can see it is simple really, just that mental block!!
    It is easy without the additional 'k' lol

    Thanks again

    James

    P.S A very useful forum, no doubt I will be posting time and again...
     
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