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Thread: induction

  1. #1
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    induction

    Looking for some help with this problem, Thanks ahead of time



    Write out an induction proof of the following equation for all n N.
    5 + 9 + 13 + ... + (4n + 5) = 2n^2 + 7n + 5.
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  2. #2
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    Base case: $\displaystyle n=1,$ so $\displaystyle \sum\limits_{k\,=\,0}^{1}{(4k+5)}=5+9=14$ besides $\displaystyle 2\cdot 1^{2}+7\cdot 1+5=14.$ Now asume valid the proposition for some $\displaystyle n=m,$ our I.H. will be $\displaystyle \sum\limits_{k\,=\,0}^{m}{(4k+5)}=2m^{2}+7m+5.$ It remains to prove that $\displaystyle \sum\limits_{k\,=\,0}^{m+1}{(4k+5)}=2(m+1)^{2}+7(m +1)+5.$ Observe that:

    $\displaystyle \begin{aligned}\sum\limits_{k\,=\,0}^{m+1}{(4k+5)} &=\underbrace{\sum\limits_{k\,=\,0}^{m}{(4k+5)}}_{ \text{I}\text{.H}\text{.}}+\,4(m+1)+5 \\
    & =2m^{2}+7m+5+4(m+1)+5 \\
    & =2(m+1)^{2}+7(m+1)+5.\quad\blacksquare
    \end{aligned}$
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