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I tried to rewrite it as $\sum_{k=1}^{30} k(30-\sum_{k=1}^{30}k)$ and then replace the $\sum_{k=1}^{30} k$ with $\frac{n(n+1)}{2}$ then substitute $n=30$ into the equation, however I am not getting the right answer.

Any help on how to solve this would be much appreciated.

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    $\begingroup$ The sum doesn't distribute like that; you should have $$30 \sum_{k = 1}^{30} k - \sum_{k = 1}^{30} k^2$$ $\endgroup$
    – user61527
    Commented Mar 23, 2014 at 18:13
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    $\begingroup$ Hint: The summation symbol does not distribute with respect to the minus sign. The summation symbol is not another factor. However, you can rewrite $k(30-k)$ as $30k-k^2$, and then you can decompose your summation into two summations, and you can apply your formula to the first summation, and look for another formula to solve the second summation. $\endgroup$ Commented Mar 23, 2014 at 18:15
  • $\begingroup$ See Faulhaber's formula. $\endgroup$
    – Lucian
    Commented Mar 23, 2014 at 19:21
  • $\begingroup$ WHO IS UPVOTING THESE? $\endgroup$
    – Alec Teal
    Commented Mar 23, 2014 at 22:00

3 Answers 3

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Hint

These two equalities are useful:

$$\sum_{k=1}^nk=\frac{n(n+1)}2$$ and $$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$$

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You can't rearrange the sum like that. Instead you should write it as

$$\sum_{k=1}^{30}(30k-k^2) = \sum_{k=1}^{30}30k-\sum_{k=1}^{30}k^2.$$

From here, employ the expressions given by Sami.

As an aside, note that your summand is very symmetric. You can rewrite it as

$$\sum_{k=1}^{14}2k(30-k)+15^2.$$

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For the sake of symmetry, let us sum from $k=0$ to $k=29$. By completing the square, we can see that we want $$\sum_1^{29}\left(15^2-(15-k)^2\right).$$ This is equal to $$(29)(15^2) -2\sum_{j=1}^{14} j^2.$$

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