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Math Help - Difference method

  1. #1
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    Difference method

    By expanding (2r+1)^5-(2r-1)^5, derive a formula for \sum{r^{4}} from 1 to n.

    I've simplified the equation to 2(80r^4+40r^2+1), where (2r+1)^5-(2r-1)^5 is f(n+1)-f(n-1). I have no slightest inclination on where to continue. anyone care to pint me in the right direction?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by smmxwell View Post
    By expanding (2r+1)^5-(2r-1)^5, derive a formula for \sum{r^{4}} from 1 to n.

    I've simplified the equation to 2(80r^4+40r^2+1), where (2r+1)^5-(2r-1)^5 is f(n+1)-f(n-1). I have no slightest inclination on where to continue. anyone care to pint me in the right direction?
    This question requires you to use the technique of telescoping sums.

    Put:

    f(r)=(2r+1)^5

    then

     f(r-1)=(2r-1)^5

    So:

    \sum_1^k [160r^4+80r^2+2]=\sum_1^k [f(r)-f(r-1)]=f(k)-f(0)

    Hence:

    160 \sum_1^k r^4=f(k)-f(0)-\left[\sum_1^k 80r^2\right] - 2k

    CB
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