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    (Idk how to do summation notation with LaTex)
    Prove that the some of integers 1 to n in the function r(r+1) = \dfrac{1}{3}n(n+1)(n+2)
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    So what I've done is show that it is true for n = 1, then I assumed that it is true for n = k and now I look at n = k + 1.
    So: The sum of all integers in the function from 1 to k + 1 = \dfrac{k}{3}(k+1)(k+2)+(k+1)(k+2  )

    So that's all the easy part and I know I need to now get it to look like
    \dfrac{k+1}{3}(k+2)(k+3)
    But I am struggling to connect the dots.
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    (Original post by Retsek)
    (Idk how to do summation notation with LaTex)
    Prove that the some of integers 1 to n in the function r(r+1) = \dfrac{1}{3}n(n+1)(n+2)
    __________________
    So what I've done is show that it is true for n = 1, then I assumed that it is true for n = k and now I look at n = k + 1.
    So: The sum of all integers in the function from 1 to k + 1 = \dfrac{k}{3}(k+1)(k+2)+(k+1)(k+2  )

    So that's all the easy part and I know I need to now get it to look like
    \dfrac{k+1}{3}(k+2)(k+3)
    But I am struggling to connect the dots.
    You have two terms with common factors (k+1)(k+2) so you need to "take them outside the brackets" to give

    (k+1)(k+2)\left[ \ \ \ ... \ \ \ \right]

    Can you continue? Factorising like this is a very common thing you need to do in induction questions.

    Also, \displaystyle \sum_{r=1}^{n} in LaTeX produces

    \displaystyle \sum_{r=1}^{n}
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    (Original post by Notnek)
    You have two terms with common factors (k+1)(k+2) so you need to "take them outside the brackets" to give

    (k+1)(k+2)\left[ \ \ \ ... \ \ \ \right]

    Can you continue? Factorising like this is a very common thing you need to do in induction questions.

    Also, \displaystyle \sum_{r=1}^{n} in LaTeX produces

    \displaystyle \sum_{r=1}^{n}
    So I managed to get \dfrac{k+3}{3}(k+1)(k+2) minutes after making this post but I didn't realise for a solid ten minutes that I could just move the 1/3 to the k+1 :erm:
 
 
 
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