0
$\begingroup$

Why is $$ \sum_{k=1}^{6}\frac{2k-1}{36} = \frac{1}{18}\sum_{k=1}^{6}k-\frac{1}{6}? $$

$\endgroup$
3
$\begingroup$

There can be ambiguity when using the $\sum$ symbol: which terms are included and which aren't? To avoid confusion, it's best to use parentheses:

$$\begin{align} \sum_{k=1}^{6}\frac{2k-1}{36} &= \sum_{k=1}^{6}\left(\frac{2k}{36}-\frac1{36}\right)\\ &=\left(\sum_{k=1}^{6}\frac{k}{18}\right) - \left(\sum_{k=1}^{6}\frac1{36}\right)\\ &=\frac1{18}\left(\sum_{k=1}^{6}k\right) - 6\left(\frac1{36}\right)\\ &=\frac1{18}\left(\sum_{k=1}^{6}k\right) - \frac16\\ \end{align} $$

$\endgroup$
  • $\begingroup$ Oh. You are right. They probably meant that. $\endgroup$ – randomgirl Apr 17 at 16:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.