Results 1 to 13 of 13
Like Tree7Thanks
  • 1 Post By JeffM
  • 3 Post By johnsomeone
  • 2 Post By johng
  • 1 Post By Prove It

Thread: mathematical induction

  1. #1
    Newbie
    Joined
    Sep 2015
    From
    europe
    Posts
    5

    mathematical induction

    mathematical induction-j.png

    Hi!!! can anyone help me with these mathematical inductions? normally, inductions are given to prove. but these are different. need HELP!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,304
    Thanks
    799
    Awards
    1

    Re: mathematical induction

    Quote Originally Posted by jkjk View Post
    Click image for larger version. 

Name:	j.PNG 
Views:	1 
Size:	7.1 KB 
ID:	34018

    Hi!!! can anyone help me with these mathematical inductions? normally, inductions are given to prove. but these are different. need HELP!!!
    I think the idea is to try to find the formulas yourself. For instance, the first few examples would be
    $\displaystyle \sum_{k = 1}^{2} (2k - 1) = (2(1) -1) + (2(2) - 1) = 1 + 3 = 4$

    $\displaystyle \sum_{k = 2}^{4} (2k - 1) = (2(2) -1) + (2(3) - 1) + (2(4) - 1) = 3 + 5 + 7 = 15$

    $\displaystyle \sum_{k = 3}^{6} (2k - 1) = (2(3) -1) + (2(4) - 1) + (2(5) - 1) + (2(6) - 1)$ $\displaystyle = 5 + 7 + 9 + 11 = 32$

    You may not be able to see the series from this but I'll give you a hint: $\displaystyle S(n) = an^2 + bn + c$. You've got three equations and three unknowns.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2015
    From
    europe
    Posts
    5

    Re: mathematical induction

    the answer is 3n^2 +2n-1 .. but i don't get it
    Last edited by jkjk; Sep 2nd 2015 at 01:34 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Feb 2014
    From
    United States
    Posts
    1,748
    Thanks
    808

    Re: mathematical induction

    It always help to look at different small values for n.

    $\displaystyle \sum_{k=1}^{2*1}(2k - 1) = \sum_{k=1}^2(2k - 1) =$

    $ \{(2 * 1) - 1\} + \{(2 * 2) - 1\} = 1 + 3 = 4 = 2^2 = (2 * 1)^2 - 0.$

    $\displaystyle \sum_{k=2}^{2*2}2k - 1) = \sum_{k=2}^4(2k - 1) =$

    $\{(2 * 2) - 1\} + \{(2 * 3) - 1\} + \{(2 * 4) - 1\} = 3 + 5 +7 = 15 = 16 - 1 = (2 * 2)^2 - 1.$

    $\displaystyle \sum_{k=3}^{2*3}(2k - 1) = \sum_{k=3}^6(2k - 1) =$

    $\{(2 * 3) - 1\} + \{(2 * 4) - 1\} + \{(2 * 5) - 1\} +\{(2 * 6) - 1\} = 5 + 7 + 9 + 11 = 32 = (2 * 3)^2 - 4.$

    Does this remind you of anything?

    I'll give you another hint.

    $\displaystyle \sum_{i=1}^m(2i - 1) = what?$
    Last edited by JeffM; Sep 2nd 2015 at 01:49 PM.
    Thanks from jkjk
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2015
    From
    europe
    Posts
    5

    Re: mathematical induction

    (2n(2n+1)(4n+1))/6 -(n+1)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Feb 2014
    From
    United States
    Posts
    1,748
    Thanks
    808

    Re: mathematical induction

    Quote Originally Posted by jkjk View Post
    (2n(2n+1)(4n+1))/6 -(n+1)
    Both topsquark and I calculated that if n = 3, the sum is 32.

    $\dfrac{2 * 3(2 *3 + 1)(4 * 3 + 1)}{6} - (3 + 1) = 7 * 13 - 4 = 87 \ne 32.$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2015
    From
    europe
    Posts
    5

    Re: mathematical induction

    then i am not clever enough to do that
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Feb 2014
    From
    United States
    Posts
    1,748
    Thanks
    808

    Re: mathematical induction

    Quote Originally Posted by jkjk View Post
    then i am not clever enough to do that
    Not clever enough to do what? He and I just did some arithmetic first. Nothing clever about that.

    Then topsquark, who is cleverer than I, saw that a quadratic would do the trick without any apparent intermediate steps. But I saw that this sum is related to the sum of the first p odd numbers and so started looking for squares too.

    He gave you an analytic approach. I gave you a pattern recognition approach. What is it that you are not seeing?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    1,061
    Thanks
    410

    Re: mathematical induction

    $\displaystyle \sum_{k = n}^{2n} (2 k - 1) = \text{?}$.

    Define $\displaystyle f(m) = \sum_{k = 1}^{m} (2 k - 1)$.

    Then $\displaystyle \sum_{k = n}^{2n} (2 k - 1) = f(2n) - f(n-1)$, because always have (for $\displaystyle n \ge 2$) that..... $\displaystyle \sum_{k = 1}^{n-1} a_k + \sum_{k = n}^{2n} a_k = \sum_{k = 1}^{2n} a_k$.

    So to find $\displaystyle \sum_{k = n}^{2n} (2 k - 1)$, it suffices to find $\displaystyle f(m)$.

    But $\displaystyle f(m) = \sum_{k = 1}^{m} (2 k - 1) = \sum_{k = 1}^{m} (2 k) + \sum_{k = 1}^{m} (- 1) = 2 \sum_{k = 1}^{m} k - \sum_{k = 1}^{m} (1)$.

    Thus $\displaystyle f(m) = 2 \left( \sum_{k = 1}^{m} k \right) - m$.

    I bet you've seen $\displaystyle \sum_{k = 1}^{m} k$ before, and if so, you can find $\displaystyle f(m)$, and so you can get a formula for $\displaystyle \sum_{k = n}^{2n} (2 k - 1) = f(2n) - f(n-1)$.
    Last edited by johnsomeone; Sep 2nd 2015 at 03:19 PM.
    Thanks from JeffM, topsquark and jkjk
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Dec 2013
    From
    Colombia
    Posts
    1,928
    Thanks
    650

    Re: mathematical induction

    I would work from the sum of the first $n$ natural numbers, $\sum_{k=1}^n k$, which, as johnsomeone says, you ought to have seen before. Given that, you have $$\sum_{k=1}^n (2k - 1) =\sum_{k=1}^{2n} k - \sum_{k=1}^n 2k$$ which just says that the sum of the first $n$ odd numbers is the sum of the first $2n$ natural numbers less the sum of the first $n$ even numbers.

    Having said that, if you work out the first few terms $\sum_{k=1}^n (2k - 1)$, you should very definitely see a pattern.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    1,144
    Thanks
    477

    Re: mathematical induction

    Hi jkjk,
    Frequently, formal manipulations with sums are helpful; what you might call "Sigma algebra".


    Thanks from JeffM and jkjk
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946

    Re: mathematical induction

    Quote Originally Posted by jkjk View Post
    Click image for larger version. 

Name:	j.PNG 
Views:	1 
Size:	7.1 KB 
ID:	34018

    Hi!!! can anyone help me with these mathematical inductions? normally, inductions are given to prove. but these are different. need HELP!!!
    You should know that the sum of the first N odd integers is N^2, so

    $\displaystyle \begin{align*} \sum_{k = 1}^{N}{ \left( 2k - 1 \right) } &= N^2 \\ \\ \sum_{k = 1}^{2n} \left( 2k - 1 \right) &= \left( 2n \right) ^2 \\ &= 4n^2 \\ \\ \sum_{k = 1}^{n - 1} \left( 2k - 1 \right) &= \left( n - 1 \right) ^2 \\ &= n^2 - 2n + 1 \\ \\ \sum_{ k = n}^{2n} \left( 2k - 1 \right) &= \sum_{ k = 1 }^{2n} \left( 2k - 1 \right) - \sum_{k= 1}^{n - 1} \left( 2k - 1 \right) \\ &= 4n^2- \left( n^2 - 2n + 1 \right) \\ &= 3n^2 + 2n - 1 \end{align*}$

    Now you could try induction if you want...
    Thanks from jkjk
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Sep 2015
    From
    europe
    Posts
    5

    Re: mathematical induction

    THANKS everyone..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mathematical induction
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Jul 15th 2012, 01:56 PM
  2. Replies: 10
    Last Post: Jun 29th 2010, 01:10 PM
  3. Mathematical Induction
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: Feb 28th 2010, 11:51 AM
  4. Mathematical induction
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 6th 2009, 10:03 PM
  5. Mathematical Induction
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: May 30th 2007, 04:21 PM

Search Tags


/mathhelpforum @mathhelpforum