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Thread: Induction

  1. #1
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    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.
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  2. #2
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    Hello, rugbygirl2!

    Evidently, this sequence begins with $\displaystyle n = 0$


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


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


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

    . . $\displaystyle \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 $\displaystyle S(k+1)$
    . . which is: .$\displaystyle \frac{\bigg[(k+1)+1\bigg]\bigg[2(k+1)+1\bigg]\bigg[2(k+1)+3\bigg]}{3} \;=\;\frac{(k+2)(2k+3)(2k+5)}{3}$ .[1]


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

    Factor: .$\displaystyle \frac{2k+3}{3}\,\bigg[(k+1)(2k+1) + 3(2k+3)\bigg] \;=\;\frac{2k+3}{3}\,(2k^2 + 3k + 1 + 6k + 9)
    $

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

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


    Our inductive proof is complete . . .

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  3. #3
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    Thank you so much for your help!!!!!!!!!!!!!
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