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Math Help - Calculus Homework, Tutor couldn't help, I've hit the wall

  1. #1
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    Calculus Homework, Tutor couldn't help, I've hit the wall

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    Last edited by Harrigan; December 1st 2008 at 05:57 AM.
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  2. #2
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    The sum of the 6th power identities, huh?. Well, here is my way. Years ago

    when I figured this out, I thought I had done something. But, it would be silly

    to think it was original, afterall.


    Consider (k+1)^{7}-k^{7}=7k^{6}+21k^{5}+35k^{4}+35k^{3}+21k^{2}+7k+1

    (n+1)^{7}-1=n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n

    Notice the coefficients?. They look binomial, don't they?

    \sum_{k=1}^{n}\left(7k^{6}+21k^{5}+35k^{4}+35k^{3}  +21k^{2}+7k+1\right) =n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n

    7\sum_{k=1}^{n}k^{6}+21\sum_{k=1}^{n}k^{5}+35\sum_  {k=1}^{n}k^{4}+35\sum_{k=1}^{n}k^{3}+21\sum_{k=1}^  {n}k^{2}+7\sum_{k=1}^{n}k+\sum_{k=1}^{n}1
    =n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n

    Now, plug in the respective identities for the powers. We must know them dong it this way.

    7\sum_{k=1}^{n}k^{6}+21\left(\frac{n^{2}(n+1)^{2}(  2n^{2}+2n-1)}{12}\right)+35\left(\frac{n(n+1)(6n^{3}+9n^{2}+  n-1)}{30}\right) +35\left(\frac{n^{2}(n+1)^{2}}{4}\right)+21\left(\  frac{n(n+1)(2n+1)}{6}\right)+7\left(\frac{n(n+1)}{  2}\right)+n

    =n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n

    Now, solve for \sum_{k=1}^{n}k^{6}

    It is lotsa horrible algebra, but if done correctly you should get:

    \sum_{k=1}^{n}k^{6}=\frac{1}{7}n^{7}+\frac{1}{2}n^  {6}+\frac{1}{2}n^{5}-\frac{1}{6}n^{3}+\frac{1}{42}n

    I am sure there is a simpler way, but you can do any of them this way as long as you know the previous ones. Just plod through the algebra.
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  3. #3
    Eater of Worlds
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    The series of reciprocals of primes
    . Here is a beautiful result proved by Euler in the year 1773. The result states that the series of all the reciprocals of primes diverges.

    (1/2) + (1/3) + (1/5) + (1/7) + (1/11) + (1/13) ...


    Your task is to prove this result. This is a bit of a tricky problem, so I will give you more elaborate hints. Assume to the contrary that the above series converges and arrive at a contradiction using the steps below. (Notation: p will denote a prime number)


    This can probably be found on line somewhere. The outline of Euler's proof
    of this can be found in "Euler, the Master of Us All" by William Dunham.



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