Results 1 to 8 of 8

Math Help - PROVE BY INDUCTION

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    3

    Angry PROVE BY INDUCTION

    I have a question and am unsure how to start it:

    Prove by induction that
    (sum from k=1 to n)of
    k(k+1)...(k+a) = 1/(a+2)*n(n+1)(n+2)....(n+a+1)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    Proof by induction

    Hello -

    Quote Originally Posted by poppy12345 View Post
    I have a question and am unsure how to start it:

    Prove by induction that
    (sum from k=1 to n)of
    k(k+1)...(k+a) = 1/(a+2)*n(n+1)(n+2)....(n+a+1)
    Tricky to know where to start, isn't it? Is it n that we vary, or a?

    Well, start in the usual way, by writing a proposition about n, and let a take care of itself. So:

    Let S_n = \sum_{k=1}^n k(k+1)...(k+a)

    Then the propositional function P(n) is defined as:

    P(n)\equiv S_n = \frac{1}{a+2}n(n+1)(n+2)...(n+a+1)

    Then write down the sum S_{n+1} as S_n + the term you get when k=(n+1).

    You'll find you can then take out lots of common factors, and eventually express this as \sum_{k=1}^{n+1} k(k+1)...(k+a), thus showing that P(n)\implies P(n+1).

    Finally, you'll need to prove that P(1) is true for any a.

    I hope I've given you enough to go on. Let me know if you need more help.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2007
    Posts
    329
    Suppose \sum_{k=1}^n\prod_{i=0}^a(k+i)=\frac 1{a+2}\prod_{j=0}^{a+1}(n+j). So we need to show that \sum_{k=1}^{n+1}\prod_{i=0}^a(k+i)=\frac 1{a+2}\prod_{j=0}^{a+1}(n+1+j).
    \sum_{k=1}^{n+1}\prod_{i=0}^a(k+i)=\sum_{k=1}^n\pr  od_{i=0}^a(k+i)+\prod_{i=0}^a(n+1+i)=\frac 1{a+2}\prod_{j=0}^{a+1}(n+j)+\prod_{i=0}^a(n+1+i) by the assumption. So this should be equal to \frac 1{a+2}\prod_{j=0}^{a+1}(n+1+j): \frac 1{a+2}\prod_{j=0}^{a+1}(n+j)+\prod_{i=0}^a(n+1+i)=  \frac 1{a+2}\prod_{j=0}^{a+1}(n+1+j)\leftrightarrow \frac 1{a+2}\prod_{j=0}^{a+1}(n+j)=\frac 1{a+2}\prod_{j=0}^{a+1}(n+1+j)-\prod_{i=0}^a(n+1+i)= \left[\prod_{i=0}^a\left(n+1+i\right)\right]\cdot\left(\frac 1{a+2}\cdot \left(n+1+(a+1)\right)-1\right)\leftrightarrow \prod_{j=0}^{a+1}(n+j)=\left[\prod_{i=0}^a\left(n+1+i\right)\right]\cdot n \leftrightarrow n\cdot (n+1)\cdot\ldots\cdot (n+a+1)=\left[(n+1)\cdot (n+2)\cdot\ldots\cdot (n+a+1)\right]\cdot n QED
    Last edited by james_bond; December 10th 2008 at 09:56 AM. Reason: Correction
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2008
    Posts
    68

    lol

    DR Duncans class by any chance?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2008
    Posts
    3
    yea
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2008
    Posts
    68
    are you still stuck on that or anything? i handed in today, pretty confident everythings right
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2008
    Posts
    3
    well i think ive managed but going tomorro. never very confident but neva mind!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Nov 2008
    Posts
    68
    im always happy to help if ya stuck... just shout, whats ur name?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. prove by induction
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 23rd 2011, 08:43 AM
  2. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  3. Prove by induction
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 10th 2009, 08:11 AM
  4. Prove induction
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 23rd 2008, 07:05 AM
  5. Prove by Induction
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 12th 2007, 10:31 AM

Search Tags


/mathhelpforum @mathhelpforum