The question is to prove (or disprove) by mathematical induction the following statement:
For $ n, a, r, \in \mathbb Z,$ and $n, a, r \in Z_{>0},$ and $r ≥ 2$ $$S(n) = \sum_{i=0}^n ar^{i-1} = an$$
The sample answer given by the book is this
$$S(2) = \sum_{i=0}^2 ar^{i-1} = a + ar = a(r+1)$$ $$≥ 3a\ (\text{ because }\ r≥2)$$ $$> 2a,\ S(2)\text{ is not true }$$
I don't understand how $S(2) = a + ar = a(r + 1)$?
Shouldn't it be $S(2) = a(2)$?
The proposed formula is actually false. Take $r=n$ (this would be a positive integer, so its good) and you will find that for $n>1$,
$$\sum_{i=0}^nar^{i-1}=\frac an+a+\color{blue}{an}+\dots+an^{n-1}>an$$
First we assume that the lower index is $i=1$ otherwise the question (and answer provided) doesn't quite make sense.
In this case, we are asked to prove that
$$S(n)=\sum_{i=1}^n ar^{i-1}=an$$ where $n,a,r\in \mathbb Z^+$ and $r\ge 2$.
This is basically the sum of a geometric progression (GP) and the closed form is clearly false for the general case unless $r=1$. As $r\ge 2$, hence the proposed statement is false.
The solution provided puts $n=2$ and evaluates the summation term by term (only $2$, so easily done by hand), i.e. $$S(2)=\sum_{i=1}^2 ar^{i-1}=a+ar=a(r+1)\ge 3a\quad \text{(as }r\ge 2) \neq 2a $$ Hence proposition is false for $n=2$.
The correct solution, for $r\ge 2$ is $$S(n)=\sum_{i=1}^n ar^{i-1}=\frac {a(r^{n-1}-1)}{r-1}$$
a + ar = a(r+1)come from? – Chan Jing Hong Jul 16 '17 at 10:46