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Math Help - Mathematical Induction

  1. #1
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    Mathematical Induction

    Hi everyone..i am struggling with this question..Any help would be appreciated

    Prove by mathematical induction that



    [img=http://img220.imageshack.us/img220/1937/formula.th.png]
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by stakr View Post
    Hi everyone..i am struggling with this question..Any help would be appreciated

    Prove by mathematical induction that



    [img=http://img220.imageshack.us/img220/1937/formula.th.png]
    First show that the basis holds:

    P\!\left(1\right)\,:\,\sum_{k=1}^1\left(3k^2+3k+1\  right)=7=8-1=\left(1+1\right)^3-1

    Now assume that \sum_{k=1}^n\left(3k^2+3k+1\right)=\left(n+1\right  )^3-1 is true for some n\in\mathbb{N}

    Now,
    \begin{aligned}\sum_{k=1}^{n+1}\left(3k^2+3k+1\rig  ht)&=\sum_{k=1}^n\left(3k^2+3k+1\right)+3\left(n+1  \right)^2+3\left(n+1\right)+1\\ &= \left(n+1\right)^3-1+3\left(n+1\right)^2+3\left(n+1\right)+1\\&=n^3+6  n^2+12n+8-1\\&=\left(n^32^0+3\cdot n^22^1+3\cdot n^12^2+n^02^3\right)-1\\&=\left(n+2\right)^3-1\end{aligned}

    This completes the induction step.

    (The second to last line is a result of the binomial theorem for cubes: (x+y)^3=x^3y^0+3x^2y^1+3x^1y^2+x^0y^3)



    Does this make sense?
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