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    Prove by the method of differences that
    \displaystyle\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1) for n>1
    Anyone can guide me on how to do this?
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    (Original post by JustDynamite)
    Prove by the method of differences that
    \displaystyle\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1) for n>1
    Anyone can guide me on how to do this?
    Is this a real question? I can't see what the 'differences' would be.
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    Maybe you just have to find someway of splitting it up such as  r(r+1)^2 -  (r-1)r^2 and then finding the sum of the series  \sum^{n}_{r=1} \left( r(r+1)^2 -(r-1) r^2 \right ) and then rearranging it.
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    The method of differences is to look at the
    sum to (k+1) - sum to k

    1/6 (k+1)(k+2)(2(k+1)+1) - 1/6 k(k+1)(2K+1)
    = 1/6 (k+1) [(k+2)(2k+3) - k(2k+1)]
    = 1/6 (k+1)[(2k^2+7k+6 -2k^2 -k]
    = 1/6 (k+1) [6k+6]
    = (k+1)^2

    hope this is reasonably clear.
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    (Original post by JustDynamite)
    Prove by the method of differences that
    \displaystyle\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1) for n>1
    Anyone can guide me on how to do this?
    (Original post by SeanFM)
    Is this a real question? I can't see what the 'differences' would be.
    That's correct Sean.

    Below is the full question:

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    (Original post by Zacken)
    That's correct Sean.

    Below is the full question:

    :lol: well remembered.

    JustDynamite, look at the link between question.
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    (Original post by MathsSir)
    The method of differences is to look at the
    sum to (k+1) - sum to k

    1/6 (k+1)(k+2)(2(k+1)+1) - 1/6 k(k+1)(2K+1)
    = 1/6 (k+1) [(k+2)(2k+3) - k(2k+1)]
    = 1/6 (k+1)[(2k^2+7k+6 -2k^2 -k]
    = 1/6 (k+1) [6k+6]
    = (k+1)^2

    hope this is reasonably clear.
    ... that's not the method of differences, pretty sure that's proof by induction :/
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    Ah thanks guys! It was question 24 from here: https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

    Looks like the top part was left out, was able to do it using the part Zacken shown
 
 
 
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