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Summing identies series help

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    Given that the sum of R^2 =1/6n (n+1)(2n+1) use the identity r^4-(r-1)^4=4r^3-6r^2+4r-1 to find the sum of the series r^3 for values 1 to n

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    (Original post by Tsrstudy)
    Given that the sum of R^2 =1/6n (n+1)(2n+1) use the identity r^4-(r-1)^4=4r^3-6r^2+4r-1 to find the sum of the series r^3 for values 1 to n

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    Use method of differences on \displaystyle \sum_{r=1}^n (r^4-(r-1)^4) and get it in terms of n.

    Then equate this result to \displaystyle \sum_{r=1}^n (4r^3-6r^2+4r-1) which you can split into 4 sums, 3 of which you can express in terms of n except the cube of course. Do that and rearrange for \displaystyle \sum_{r=1}^n (r^3)
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    (Original post by RDKGames)
    Use method of differences on \displaystyle \sum_{r=1}^n (r^4-(r-1)^4) and get it in terms of n.

    Then equate this result to \displaystyle \sum_{r=1}^n (4r^3-6r^2+4r-1) which you can split into 4 sums, 3 of which you can express in terms of n except the cube of course. Do that and rearrange for \displaystyle \sum_{r=1}^n (r^3)
    Ah of course , thank you so much this was a great help!
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    (Original post by Tsrstudy)
    Ah of course , thank you so much this was a great help!
    my issue is am i allowed to sum in the value for r which is 1/2n(n+1), im not sure you are since u are specifically only given the sum of the series r^2, if you are allowed well then the problem is resolved
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    (Original post by Tsrstudy)
    my issue is am i allowed to sum in the value for r which is 1/2n(n+1), im not sure you are since u are specifically only given the sum of the series r^2, if you are allowed well then the problem is resolved
    I'm sure you're allowed, though I do not know the context of this question as in where it comes from and what the restrictions usually are. You can always derive that result fairly quickly as well if required.
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    (Original post by RDKGames)
    I'm sure you're allowed, though I do not know the context of this question as in where it comes from and what the restrictions usually are. You can always derive that result fairly quickly as well if required.
    Ok, thanks very much for your time
 
 
 
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