# Evaluating $\sum_{k=1}^{30} k(30-k)$

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.

• The sum doesn't distribute like that; you should have $$30 \sum_{k = 1}^{30} k - \sum_{k = 1}^{30} k^2$$
– user61527
Commented Mar 23, 2014 at 18:13
• 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. Commented Mar 23, 2014 at 18:15
• Commented Mar 23, 2014 at 19:21
• WHO IS UPVOTING THESE? Commented Mar 23, 2014 at 22:00

Hint

These two equalities are useful:

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

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

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

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 $$(29)(15^2) -2\sum_{j=1}^{14} j^2.$$