Do not understand what is meant by "in terms of n"

Question

I took calculus 5 years ago so I am unaware of what is meant by in terms of the variables in this summation example.

specifically these two questions. please help me understand what the steps are to solving these. I was unable to google my malfunction.

5. What is the value of the following summation in terms of $n$: $\sum\limits_{=+1}^{3} (2−3)$?

6. Determine the value of this summation, in terms of $a$, $b$, $c$ and $d$. Assume that $a < b$ and that all four are positive integers. $\sum\limits_{=}^{b} (+)$.

Answer 1

When we say expression in terms of $n$, your final answer should be an expression involving $n$.

For your problem, I will help you with the first one but I will leave the second one to you.

\begin{align}\sum_{i=n+1}^{3n}(2i-3)&=2\sum_{i=n+1}^{3n}i - 3\sum_{i=n+1}^{3n}1 \\ &=2[(n+1) + (n+2) + \ldots + 3n]-3(2n)\\ &= 2 \cdot \frac{2n}{2}\left[(n+1)+3n \right]-6n\end{align}

Try to simplify the expression above. As you can see, the final answer involves $n$.

Answer 2

Hint: Each of these is the sum of consecutive terms of arithmetic sequences. As is well known, such a sum is given by the formula $$ \frac{\text{first term}+\text{last term}}2\times\text{number of terms}. $$

Answer 3

To compute the second sum, you can write $\sum_{i=a}^b (ci + d) = c(\sum_{i=1}^b i - \sum_{i=1}^{a-1} i) + d (b-a+1)$ and apply the formula for the sum of consecutive integers $\sum_{k=1}^n k = n(n+1)/2$.

Answer 4

They want a function that doesn't involve a sum:

$$\sum_{i=a}^{b}(ci-d) = c\sum_{i=a}^{b}i - \sum_{i=a}^{b}d = c\sum_{i=a}^{b}i - (b-a+1)d$$

Which means you need a closed form solution for $\sum_{i=a}^{b}i$. If you expand the sum you have: $$\sum_{i=a}^{b}i = a + (a+1) + (a+2) + ... + (b-2) + (b-1) + b$$ If you group the first and last term, the second and before last and so on, you get: $$(a+b) + (a+1 + b-1) + (a+2+b-2) + ... = (a+b) + (a+b) + (a+b) + ...$$ And there are $(b-a+1)/2$ such terms, then the closed form solution is: $$\sum_{i=a}^{b}i = \frac{(a+b)(b-a+1)}{2}$$

Finally:

$$\sum_{i=a}^{b}(ci-d) = \left(\frac{ac+bc}{2}-d\right)(b-a+1)$$