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Math Help - sum of first n terms

  1. #1
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    sum of first n terms

    i got stucked at part a , how should i proceed form the previous steps? i know that the (2n-1)^2 -(2n)^2 +(2n+1)^2 = (1)^2 -(2)^2 +(3)^2 sum of first n terms-dsc_0001-2-2-.jpgsum of first n terms-dsc_0001-3-1-.jpgsum of first n terms-dsc_0002-2-1-.jpg
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  2. #2
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    Re: sum of first n terms

    just add -n(2n+1)+(2n+1)^2
    so (2n+1)(-n+2n+1)
    sum=(2n+1)(n+1)
    of a part
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  3. #3
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    Re: sum of first n terms

    Quote Originally Posted by prasum View Post
    just add -n(2n+1)+(2n+1)^2
    so (2n+1)(-n+2n+1)
    sum=(2n+1)(n+1)
    of a part
    can you explain why would you do so? i cant understand
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  4. #4
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    Re: sum of first n terms

    u know the sum of 1^2-2^2+3^2-4^2+...(2n-1)^2-(2n)^2=-n(2n+1)
    u have extra term=(2n+1)^2
    so just add these two to get ur final sum as:
    -n(2n+1)+(2n+1)^2
    so (2n+1)(-n+2n+1)
    sum=(2n+1)(n+1)
    of a part
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  5. #5
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    Re: sum of first n terms

    why is it not necessary to find the new sum again? but just to take the old sum and plus with the new terms?
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  6. #6
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    Re: sum of first n terms

    because u have already found out from the previous question
    the sum 1^2-2^2+3^2-4^2+...(2n-1)^2-(2n)^2=-n(2n+1)
    in these type of questions u use what u have proved above
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