Results 1 to 3 of 3

Math Help - Induction

  1. #1
    Newbie
    Joined
    Nov 2008
    From
    rochester, ny
    Posts
    8

    Induction

    I need help with a mathematical induction proof to prove that
    (1^2)+(3^2)+(5^2)+.....+(2n+1)^2 = ((n+1)(2n+1)(2n+3))/3 for all integers n is greater than or equal to zero.

    I understand induction but I can't figure out the algebra, any help would be appreciated! Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,811
    Thanks
    701
    Hello, rugbygirl2!

    Evidently, this sequence begins with n = 0


    Prove: . 1^2+3^2+5^2+ \hdots + (2n+1)^2 \;=\;\frac{(n+1)(2n+1)(2n+3)}{3}
    Verify S(1)\!:\;\;1^2+3^2 \:=\:\frac{2\cdot3\cdot5}{3} . . . True!


    Assume S(k)\!:\;\;1^2+3^2+5^2 + \hdots + (2k+1)^2 \;=\;\frac{(k+1)(2k+1)(2k+3)}{3}


    Add (2k+3)^2 to both sides:

    . . \underbrace{1^2 + 3^2 + 5^2 + \hdots + (2k+3)^2}_{\text{The left side of }S(k+1)} \;=\;\frac{(k+1)(2k+1)(2k+3)}{3} + (2k+3)^2


    We must show that the right side is the right side of S(k+1)
    . . which is: . \frac{\bigg[(k+1)+1\bigg]\bigg[2(k+1)+1\bigg]\bigg[2(k+1)+3\bigg]}{3} \;=\;\frac{(k+2)(2k+3)(2k+5)}{3} .[1]


    The right side is: . \frac{(k+1)(2k+1)(2k+3)}{3} + (2k+3)^2

    Factor: . \frac{2k+3}{3}\,\bigg[(k+1)(2k+1) + 3(2k+3)\bigg]  \;=\;\frac{2k+3}{3}\,(2k^2 + 3k + 1 + 6k + 9)<br />

    . . . . = \;\frac{2k+3}{3}\,(2k^2 + 9k+10) \;=\;\frac{2k+3}{3}\,(k+2)(2k+5)

    . . . . = \;\frac{(k+2)(2k+3)(2k+5)}{3}\quad\hdots . which equals [1]


    Our inductive proof is complete . . .

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2008
    From
    rochester, ny
    Posts
    8
    Thank you so much for your help!!!!!!!!!!!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Strong induction vs. structural induction?
    Posted in the Discrete Math Forum
    Replies: 13
    Last Post: April 21st 2011, 12:36 AM
  2. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  3. induction help
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: April 19th 2010, 05:39 AM
  4. Mathemtical Induction Proof (Stuck on induction)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 8th 2009, 09:33 PM
  5. Induction!
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 7th 2008, 04:10 PM

Search Tags


/mathhelpforum @mathhelpforum