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Tough series question (well for me anyway! [FP1]) Watch

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    See the attachment .

    Stuck on part b).

    Thanks
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    (Original post by marek35)
    See the attachment .

    Stuck on part b).

    Thanks
    did you get to this stage?

    kn(n+1)(an^2+bn+c)=8\frac{n}{6}(  n+1)(2n+1)
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    Yeah I did i simplified that to-> 1/4n(n+1) (n^2+n+2)= 4/3 n (n+1)(2n+1).

    Then I didnt know what to do after that

    cheers
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    (Original post by Pheylan)
    did you get to this stage?

    kn(n+1)(an^2+bn+c)=8\frac{n}{6}(  n+1)(2n+1)
    Yeah I did i simplified that to-> 1/4n(n+1) (n^2+n+2)= 4/3 n (n+1)(2n+1).

    Then I didnt know what to do after that

    cheers
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    (Original post by marek35)
    Yeah I did i simplified that to-> 1/4n(n+1) (n^2+n+2)= 4/3 n (n+1)(2n+1).

    Then I didnt know what to do after that

    cheers
    assuming your values of a, b, c and k are right

    get rid of the fractions (no one likes a fraction):

    3n(n+1)(n^2+n+2)=16n(n+1)(2n+1)

    take out a factor of n(n+1):

    n(n+1)[3(n^2+n+2)-16(2n+1)]=0

    simplify:

    n(n+1)(3n^2-29n-10)=0

    can you finish it off?
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    (Original post by Pheylan)
    assuming your values of a, b, c and k are right

    get rid of the fractions (no one likes a fraction):

    3n(n+1)(n^2+n+2)=16n(n+1)(2n+1)

    take out a factor of n(n+1):

    n(n+1)[3(n^2+n+2)-16(2n+1)]=0

    simplify:

    n(n+1)(3n^2-29n-10)=0

    can you finish it off?
    Yeah done it solved it and got n=10

    cheers
 
 
 
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