# Summation and brackets

Is it possible to write $\sum_{i=1}^k y_i\log p_i +(n_i-y_i)\log(1-p_i)+\log \binom{n_i}{y_i}$ as one mathematician said it is correct but another said that one should write $\sum_{i=1}^k\left ( y_i\log p_i +(n_i-y_i)\log(1-p_i)+\log \binom{n_i}{y_i} \right )$

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To expand on @vonbrand’s answer, the fact that the index $i$ appears in all three terms almost guarantees that they all belong to the summation, but you shouldn’t make the reader work that hard: it’s much better to enclose the entire summand in parentheses, so that it’s immediately obvious what is being summed. –  Brian M. Scott Apr 19 '13 at 9:46

Expressions like $$\tag?\sum_{i=1}^n a_i + 1$$ have two possible interpretations: $$\tag1\sum_{i=1}^n (a_i + 1)$$ or $$\tag2\left(\sum_{i=1}^n a_i\right) + 1.$$ Among others because of the uglyness of $(2)$ it is customary to interprete $(?)$ as $(2)$ and use explicit parentheses if one wants to have $(1)$. Note however, that no parentheses are necessary/customary for $$\sum_{i=1}^n a_i\cdot 2=\sum_{i=1}^n (a_i\cdot 2)= \left(\sum_{i=1}^n a_i\right)\cdot 2=2\sum_{i=1}^n a_i$$
The only thing in $$\sum_{i=1}^k y_i\log p_i +(n_i-y_i)\log(1-p_i)+\log \binom{n_i}{y_i}$$ that tells us that the summation should involve all terms is that all terms have an index $_i$ on them. For example, the following example is much unclearer: $$\sum_{i=1}^n a_i+b$$ which could either mean $$\left(\sum_{i=1}^n a_i\right)+b\quad\text{or}\quad\sum_{i=1}^n \left(a_i+b\right)=\left(\sum_{i=1}^n a_i\right)+nb,$$ which is two completely different things. So, it's a good habit to include parentheses as @vonbrand also answers.