I stumbled upon two problems that I solved in an exam a while ago; however, the only reason why I solved them was through trial and error by picking each choice from the choices available, I set an upper limit and compared the two values from the closed form and the summation, and when they matched I knew that that closed form was the right answer. But this isn't very beneficial since I don't know the logic behind converting the summation to a closed form. So my question is, what are the steps that converted these two summations to its respective closed forms
3 Answers
$\begingroup$The quicker way is to use arithmetic series directly but I am showing you a more fundamental approach.
I will show you the first question and leave the second one as practice.
Result needed:
$$\sum_{i=1}^n 1 = n$$ Since we just counting how many terms are there. $$\sum_{i=1}^n i=\frac{n}{2}(1+n)$$ since this is just the arithmetic series.
\begin{align} \sum_{i=0}^n (5-i) &= 5 \sum_{i=0}^n1 - \sum_{i=0}^n i \\ &=5(n+1)-\sum_{i=1}^n i \end{align}
What remains are just algebraic manipulations.
$\endgroup$ 0 $\begingroup$Use the formula of the summation of arithmetic series
$$\dfrac{m(A+L)}2$$ where $m$ is the number of terms, $A,L$ being the first & last terms respectively
$\endgroup$ 2 $\begingroup$It's often easier to guess than calculate given a multiple choice question, and it can be a lot easier to guess if you choose the right cases to check. For question 1. for example, the sum for $n=0$ is $\sum_{i=0}^{0}(5-i)=5-0=5\,$. The only answer which evaluates to $5$ for $n=0$ is A.
To derive the closed form, it's enough to remember that $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}\,$, then for example:
$$\sum_{i=0}^{n}(5-i)=\left(5 \, \sum_{i=0}^{n} 1\right) - \left(\sum_{i=0}^{n} i\right) = 5 (n+1) - \left( 0 + \sum_{\color{red}{i=1}}^{n} i \right)= 5(n+1) - \frac{n(n+1)}{2} = \cdots$$
$\endgroup$
