In complexity theory, it is sometimes necessary to find the "closed-form solution" of a summation.
This was put in our exam guide as "solving arithmetic and geometric series", which I initially understood as finding the nth term and the sum to n of a series.
I now understand that it refers to finding the closed-form solution of the sum and I know that the formulae for these three things are different.
What is the relation between finding the nth term and the sum, and finding the closed-form solution, if any?
It is easiest to see the differences using an example, so consider the geometric series $\sum_{k=1}^{j}{ar^{k-1}}$ with initial value $a$ and common ratio $r$.
To find the nth term evaluate the summand $ar^{k-1}$ when $k = n$.
1st term = $ar^{1-1} = ar^0 = a$
2nd term = $ar^{2-1} = ar^1 = ar$
(n-1)th term = $ar^{(n-1)-1} = ar^{n-2}$
nth term = $ar^{n-1}$
To find the closed form, rewrite the sum so that it does not contain a summation. Essentially, it is a formula for the sum of the first j terms of the summation which can be evaluated in one step, instead of j steps. We do this by factoring out the initial value then multiplying both sides by $(r - 1)$ which will cancel all but the last term.
$\sum_{k=1}^{j}{ar^{k-1}} = a + ar + ar^2 + \dots + ar^{j-2} + ar^{j-1}$
$\sum_{k=1}^{j}{ar^{k-1}} = a(1 + r + r^2 + \dots + r^{j-2} + r^{j-1})$
$(r -1)\sum_{k=1}^{j}{ar^{k-1}} = a(r - 1)(1 + r + r^2 + \dots + r^{j-2} + r^{j-1})$
$(r - 1)\sum_{k=1}^{j}{ar^{k-1}} = a(r + r^2 + r^3 + \dots + r^{j-1} + r^{j} - 1 - r - r^2 - \dots - r^{j-2} - r^{j-1})$
$(r - 1)\sum_{k=1}^{j}{ar^{k-1}} = a(r^{j} - 1)$
$\sum_{k=1}^{j}{ar^{k-1}} = a\frac{r^{j} - 1}{r - 1}$
So $a\frac{r^{j} - 1}{r - 1}$ is the closed form of the geometric series.
Finally, we can find the sum of the first j terms using either the summation or its closed form and get the same answer, e.g. for $a = 3, r = \frac{1}{2}, j = 4$:
$\sum_{k=1}^{4}{3\left(\frac{1}{2}\right)^{k-1}} = 3 + 3\left(\frac{1}{2}\right) + 3\left(\frac{1}{4}\right) + 3\left(\frac{1}{8}\right) = 3\left(\frac{15}{8}\right)$
$3\left(\frac{\left(\frac{1}{2}\right)^{4} - 1}{\frac{1}{2} - 1}\right) = 3\left(\frac{\frac{1}{16} - 1}{-\frac{1}{2}}\right) = 3\left(\frac{-\frac{15}{16}}{-\frac{8}{16}}\right) = 3\left(\frac{15}{8}\right)$