Don't understand an induction solution.

Question

I tried to solve the following induction proof question before giving up and looking at the solution, however the solution didn't really help me either.

Question

Re-prove the formula for the sum of a geometric series with first term $a$, common ratio $r \neq 1$ and $n$ terms, by induction on $n$.

Solution

Let $P(n)$ be the statement "$a+ar+...+ar^{n-1}=\dfrac{a(r^n-1)}{(r-1)^n}. (r \neq 1, \in \mathbb{N})$.

Base Step: $a=\dfrac{a(r-1)}{(r-1)}$; so $P(1)$ is true.

Induction Step: Assume $P(k)$ is true. Then

$(1): a+ar+...ar^{k-1}+ar^k =\dfrac{a(r^k-1)}{r-1}+ar^k$, (by induction hypothesis)

$ (2): =\dfrac{a((r^k-1)+r^k(r-1))}{r-1}$

$ (3):=\dfrac{a(r^{k+1}-1)}{r-1}.$

So $P(k+1)$ is true. Hence by induction, $P(n)$ is true for all $n \in \mathbb{N}$.

Problems

$(i)$ So, in line $(1)$, I don't under why only $(r-1)$ is in the denominator when it's $(r-1)^n$ in the original statement. I was under the assumption that when assuming $P(k)$, you just change all the $n$ to $k$, why is this not necessarily true?

$(ii)$ How do you get from step $(2)$ to step $(3)$? It seems quite obscure.

Thanks.

Answer 1

(i) That's just a typo here, what is meant and used later on is $$ \tag 1 P(n): \qquad \sum_{i=0}^{n-1} ar^i = a \frac{r^n- 1}{r-1} $$ (ii) To go from (2) to (3), lets have a look at the numerator of (2), we have $$ a\bigl(r^k - 1 + r^k(r-1)\bigr) = a(r^k - 1 +r^{k+1} - r^k) = a(r^{k+1} - 1) $$ hence $$ a\frac{r^k - 1 + r^k(r-1)}{r-1} = a\frac{r^{k+1} - 1}{r-1} $$