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Proof by induction

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove by induction that the sum of the first n "squares" is n(n+1)(2n+1)/6
    In other words prove ([itex]\forall[/itex]n) P(n) where P (n) is [itex]\sum[/itex][itex]^{n}_{i=1}[/itex]¡[itex]^{2}[/itex]=[itex]\frac{n(n+1)(2n+1)}{6}[/itex]

    2. Relevant equations
    This is just not clicking for me right now. I have no idea if i am just epic failing at factoring or doing something else wrong.

    3. The attempt at a solution
    I did the base case and proved P (1) true easily enough.
    Induction case: suppose P (k), then [itex]\sum[/itex][itex]^{k}_{i=1}[/itex]i[itex]^{2}[/itex]=[itex]\frac{k (k+1)(2k+1)}{6}[/itex] show P (k+1) is true, then ∑[itex]^{k+1}_{i=1}[/itex]i[itex]^{2}[/itex]=[itex]\frac{(k+1)(k+1)(2k+1)}{6}[/itex]

    =[itex]\frac{k (k+1)(2k+1)}{6}[/itex] + (k+1)[itex]^{2}[/itex]

    Now i know i need to get itto work out to the [itex]\frac{(k+1)(k+1)(2k+3)}{6}[/itex] but i cant manage to get itto work out at all.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 3, 2014 #2


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    You mean you need to work it out to ##\frac{(k+1)(k+\color{red}2)(2k+3)}{6}##. You just need to continue with the algebra. Add up those two terms and simplify it by factoring out ##k+1##.
  4. Mar 4, 2014 #3
    Oh gees ya thats probably what my problem was, thank you sometimes you just need another set of eyes.
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