Results 1 to 5 of 5

Math Help - Proof By Induction

  1. #1
    Newbie kaylakutie's Avatar
    Joined
    Jan 2010
    Posts
    16

    Proof By Induction

    For my Linear Algebra class we have to do Proof By Induction problems. I got #1 and #2 on the first page and I got the whole second page. But what do you do on #3 and #4 on the first page? Images attached. Thanks you for your time - I appreciate it.
    Attached Thumbnails Attached Thumbnails Proof By Induction-q1.jpg   Proof By Induction-q2.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie kaylakutie's Avatar
    Joined
    Jan 2010
    Posts
    16

    Problem 1

    <br />
\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}<br />

    Base Case

    <br />
\sum_{i=1}^{1} i^2 = 1^2 = 1 , \frac{1(1+1)(2*1+1)}{6} = \frac{6}{6} = 1<br />

    Assume k \in S

    <br />
\sum_{i=1}^{k} i^2 = \frac{n(n+1)(2n+1)}{6}<br />


    Prove k+1 \in S

    <br />
\sum_{i=1}^{k+1} i^2 = \frac{(k+1)(k+2)(2k+3)}{6} = \frac{(k+1)(2k^2+7K+6)}{6} = \frac{2k^3+9k^2+13k+6}{6}<br />
    <br />
\sum_{i=1}^{k+1} i^2 = \sum_{i=1}^{k} i^2 + (k+1)^2<br />
    <br />
\sum_{i=1}^{k+1} i^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + k^2+2k+1<br />
    <br />
= \frac{(k^2+k)(2k+1)}{6} + \frac{6k^2+12k+6}{6} = \frac{2k^3+3k^2+k}{6} + \frac{6k^2+12k+6}{6}<br />
    <br />
= \frac{2k^3+9k^2+13k+6}{6}<br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    For the 3rd one,

    1!<2^1
    2!<2^2
    3!<2^2
    4!>2^4

    If n!>2^n,\ then (n+1)!>2^{n+1}

    (n+1)!=(n+1)n(n-1)(n-2)(n-3)... =(n+1)n!

    (n+1)n!>2^{n+1} ?

    (n+1)n!>(2)2^n ?

    As n+1>2 for n>2

    (n+1)!>2^{n+1} if n!>2^n
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    For 4,

    \sum_{i=0}^na^i=\frac{1-a^{n+1}}{1-a},\ a\ne1

    n=0,\ a^0=\frac{1-a}{1-a}

    n=1,\ a^0+a^1=\frac{1-a^2}{1-a}=\frac{(1-a)(1+a)}{1-a}=1+a

    If \sum_{i=0}^na^i=\frac{1-a^{n+1}}{1-a} then \sum_{i=0}^{n+1}a^i=\frac{1-a^{n+2}}{1-a}

    Proof

    \frac{1-a^{n+1}}{1-a}+a^{n+1}=\frac{1-a^{n+1}+(1-a)a^{n+1}}{1-a}

    =\frac{1-a^{n+1}+a^{n+1}-a^{n+2}}{1-a}=\frac{1-a^{n+2}}{1-a}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Finally,

    1=1
    1+3=4
    1+3+5=9
    1+3+5+7=16

    \sum_{i=1}^n2n-1=n^2

    Hence \sum_{i=1}^n2(n+1)-1 should =(n+1)^2

    \sum_{i=1}^n2n+1=n^2+2n+1 ?

    Proof

    \sum_{i=1}^n2n-1+[2(n+1)-1]=n^2+[2(n+1)-1]=n^2+2n+1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof by induction
    Posted in the Algebra Forum
    Replies: 13
    Last Post: January 31st 2011, 04:41 PM
  2. an induction proof
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: October 5th 2009, 01:31 PM
  3. proof by induction ...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 8th 2009, 02:07 PM
  4. Mathemtical Induction Proof (Stuck on induction)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 8th 2009, 09:33 PM
  5. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: June 8th 2008, 01:20 PM

Search Tags


/mathhelpforum @mathhelpforum