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    I'm stuck on the following question:

    Prove by induction that\displaystyle\sum_{r=1}^n (3r^2 -5) = \frac{1}{2}n(n+3)(2n-3) for all n \geq 1

    I've done the n=1 step, which is true as both come out as -2

    I then replaced n with k and assume it to be true.

    I then put in k+1 in with the (k+1)th term being 3(k+1)^2 - 1 however I'm not sure if it is right. If it is then I can't do the algebra proving that this is equal to the original statement with k+1

    Thanks in advance
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    Do you mean 3(k+1)^2-5 and not 3(k+1)^2-1?
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    (Original post by heymynameisben)
    I'm stuck on the following question:

    Prove by induction that\displaystyle\sum_{r=1}^n (3r^2 -5) = \frac{1}{2}n(n+3)(2n-3) for all n \geq 1

    I've done the n=1 step, which is true as both come out as -2

    I then replaced n with k and assume it to be true.

    I then put in k+1 in with the (k+1)th term being 3(k+1)^2 - 1 however I'm not sure if it is right. If it is then I can't do the algebra proving that this is equal to the original statement with k+1

    Thanks in advance
    I haven't looked at where you got that from you in detail but you should follow this kind of structure: (after the basis case obviously which you've done)
    \displaystyle\sum_{r=1}^{k+1} (3r^2 -5) =3(k+1)^2 - 5 + \displaystyle\sum_{r=1}^{k} (3r^2 -5)
    And by assuming that \displaystyle\sum_{r=1}^k (3r^2 -5) = \frac{1}{2}k(k+3)(2k-3) is true then
    \displaystyle\sum_{r=1}^{k+1} (3r^2 -5) =3(k+1)^2 - 5 + \frac{1}{2}k(k+3)(2k-3)

    Do you know where to go from there:
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    (Original post by dr_98_98)
    Do you mean 3(k+1)^2-5 and not 3(k+1)^2-1?
    Yes I did, my mistake
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    (Original post by ElMoro)
    I haven't looked at where you got that from you in detail but you should follow this kind of structure: (after the basis case obviously which you've done)
    \displaystyle\sum_{r=1}^{k+1} (3r^2 -5) =3(k+1)^2 - 5 + \displaystyle\sum_{r=1}^{k} (3r^2 -5)
    And by assuming that \displaystyle\sum_{r=1}^k (3r^2 -5) = \frac{1}{2}k(k+3)(2k-3) is true then
    \displaystyle\sum_{r=1}^{k+1} (3r^2 -5) =3(k+1)^2 - 5 + \frac{1}{2}k(k+3)(2k-3)

    Do you know where to go from there:
    That's the stage I got to but I'm afraid I'm not totally sure when to go from there.
    Do you try and prove it's equal to \frac{1}{2}(k+1)(k+4)(2k-1) ?
    If that's right then I'm having difficulties rearranging to equal each other. Thanks
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    (Original post by heymynameisben)
    That's the stage I got to but I'm afraid I'm not totally sure when to go from there.
    Do you try and prove it's equal to \frac{1}{2}(k+1)(k+4)(2k-1) ?
    If that's right then I'm having difficulties rearranging to equal each other. Thanks
    Can you post your working? Hint: factor theorem.
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    (Original post by ElMoro)
    Can you post your working? Hint: factor theorem.
    Boom, just got it. Thanks for your help
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    (Original post by heymynameisben)
    Boom, just got it. Thanks for your help
    Welcome ^_^
 
 
 
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