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Thread: Mathematical induction

  1. #1
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    Mathematical induction

    Use the principle of mathematical induction to prove the
    n
    Σ r^3 = 1/4n^2(n+1)^2.
    r=1

    I need help on this problem.
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  2. #2
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    Hi

    You can see it works for $\displaystyle n=1,$ i.e. $\displaystyle \sum\limits_{r=1}^1r^3=\left( \frac{1(1+1)}{2}\right)^2$

    Now assume it is true for some $\displaystyle n\geq 0.$

    Write:

    $\displaystyle \sum\limits_{r=1}^{n=1}r^3$
    $\displaystyle =\sum\limits_{r=1}^nr^3+(n+1)^3$
    $\displaystyle =\left( \frac{n(n+1)}{2}\right)^2+(n+1)^3$ (by induction hypothesis)
    $\displaystyle =\left( \frac{n(n+1)}{2}\right)^2+(n+1)^2(n+1)$

    Can you end?
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  3. #3
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    Thank you, but I cant end till I see. Once I see it I can solve another problem. I would appreciate if ou can end it.
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  4. #4
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    Well the idea is to factorize by $\displaystyle (n+1)^2$ which appears in the two terms:

    $\displaystyle \left( \frac{n(n+1)}{2}\right)^2+(n+1)^2(n+1)$
    $\displaystyle =(n+1)^2\left( \frac{n^2}{4}+(n+1)\right)$
    $\displaystyle =(n+1)^2\left( \frac{n^2+4n+4}{4}\right)$
    $\displaystyle =(n+1)^2\left( \frac{(n+2)^2}{4}\right)$
    $\displaystyle =\left( \frac{(n+1)(n+2)}{2}\right)^2$

    which is what you wanted.
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  5. #5
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    Smile

    Quote Originally Posted by clic-clac View Post
    Hi

    You can see it works for $\displaystyle n=1,$ i.e. $\displaystyle \sum\limits_{r=1}^1r^3=\left( \frac{1(1+1)}{2}\right)^2$

    Now assume it is true for some $\displaystyle n\geq 0.$

    Write:

    $\displaystyle \sum\limits_{r=1}^{n=1}r^3$
    $\displaystyle =\sum\limits_{r=1}^nr^3+(n+1)^3$
    $\displaystyle =\left( \frac{n(n+1)}{2}\right)^2+(n+1)^3$ (by induction hypothesis)
    $\displaystyle =\left( \frac{n(n+1)}{2}\right)^2+(n+1)^2(n+1)$

    Can you end?
    i think he should assume that it is true for $\displaystyle n\geq 1$,no ?
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  6. #6
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    Yeah indeed it would be better according to what was done before

    By the way the case $\displaystyle n=0$ is also true, but isn't very interesting (nothing more than $\displaystyle 0=0$)
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