# How do I substitute an expression into a summation term that includes a fraction?

I'm working on some economics homework and there is a problem I'm finding particularly difficult. The first question asks us to find the function that a firm will use to determine how much CO2 to emit under a cap-and-trade system, given a certain cost function for reducing its emissions. I've arrived at the expression e = (2αi-p)/2α²), where α is a constant, i is a numerical value assigned to each firm and p is the price of emitting 1 unit of CO2.

The next question is where it gets difficult. The question asks us to determine ∑(i=1, n) e, and it also tells us that ∑(i=1, n) i = N(N + 1)/2. Naturally, I substituted N(N + 1)/2 in for i, only to hit a brick wall. The TA for the course informed me that I can't directly substitute in N(N + 1)/2 for i if there is a fraction in the e-term, and advised me to "try separating the two terms so that one part has the sum i expression and the other part doesn't have i". I've been digging into the rules about summation and come up short; I also tried rearranging the equation first and then substituting for i, rather than the other way around, but I just got the same solution as when I substituted, then rearrange. Is there some summation rule I'm missing here?

Would appreciate any insight. Thanks for the attention!

• But is there REALLY a fraction here? At least one of $N$ or $N+1$ must be an even number, therefore $N(N+1)$ is divisible by $2$. Oct 8, 2022 at 2:32
• @DavidH Not to mention, $\sum i$ must be an integer as all the $i$s are integer so if $\sum i = \frac {N(N+1)}2$ then it has to be that $\frac {N(N+1)}2$ has to be an integer.... which is not a contradiction $\frac M2 \in \mathbb Z$ just means that $2|M$ and it shouldn't be any surprise that $2|N(N+1)$ because..... what David H said. Oct 8, 2022 at 3:33
• I'm not sure what you mean but substitutint $N(N+1)/2$ for $i$ nor why that'd be "natural". $\frac {N(N+1)}2 \ne i$; $\frac {N(N+1)}2 =$ the sum of all the $I$, not just one single $i$. I don't see what you are attempting to do with substituting. Oct 8, 2022 at 3:36
• The fraction is not the issue and not what the ta is talking about. What the TA wants you do do is rewrite $\sum \frac {2api-p}{2a^2}$ as $\sum (C_1\cdot i +C_2)$. If you can do that you can rewrite $\sum (C_1\cdot i +C_2)=\sum (C_1\cdot i)+\sum C_2= C_1(\sum i)+\sum C_2$. If you do that you can substitute $\sum i$ with $\frac {N(N+1)}2$. Oct 8, 2022 at 3:40

It seems that you have to calculate $$\sum\limits_{i=1}^{n} \frac{2ai-p}{2a^2}$$. The first step is factoring out the factor which does not depend on index $$i$$. This is the denominator. Thus we get $$\frac{1}{2a^2}\cdot \sum\limits_{i=1}^{n} (2ai-p)$$. Then we have two summands in the brackets. I ommit the factor $$\frac{1}{2a^2}$$ for the next steps. We can write for each summand a sigma sign.
$$\sum\limits_{i=1}^{n} 2ai-\sum\limits_{i=1}^{n}p$$
$$2 a\cdot \sum\limits_{i=1}^{n} i-p\cdot \sum\limits_{i=1}^{n}1$$
You know what the result of the first sigma sign is. And $$\sum\limits_{i=1}^{n}1$$ is just $$n$$. After some simplifications you put a bracket around the term and divide it by $$2a^2$$.