I am asked to compute $\int_1^2 \sqrt x$ from the definition of integral. From my textbook:
Choose $r= 2^{1/n} > 1$ and let $N_n$ be the grid with division points
$1 = r^0 < r <r^2< \cdots <r^{n-1} <r^{n} =2$.
The longest of the intervals determined by this grid is the last, so that $d(N_n) = 2-r^{n-1} = 2(1 - \frac{1}{r})$. Since $\lim_{ n \to \infty} r = \lim_{n \to \infty} 2^{1/n} = 1, \lim_{n \to \infty} d(N_n) = 0$, and
$S(N_n)= \sqrt{r}(r-1) + \sqrt{r^2}(r^2 - r) + \cdots + \sqrt{r^n}(r^{n}-r^{n-1})$ (1)
= $ \sqrt{r}(r-1)(1 + r\sqrt{r} + [r\sqrt{r}]^2 + \cdots + [r\sqrt{r}]^{n-1})$ (2)
= $\sqrt{r}(r-1) \frac{r^{\frac{3n}{2}}-1}{r^{\frac{3}{2}}-1}$ (3)
I wanted to know how I get from (2) to (3)?
The sum of a finite geometric series with ratio $x$ is given by
$$1 + x + x^2 + ... + x^n = \frac{x^{n+1} - 1}{x-1}$$
Notice that you have a geometric series, with ratio $r\sqrt{r} = r^{3/2}$, and it becomes clear.