Induction proof dealing with geometric series

Question

$1+r+(r^2)+...+r^n= \frac{1-r^{n+1}} {1-r}$

Any help would be appreciated in solving the geometric series.

Answer 1

Notice, the following steps

Step 1: setting $n=1$, we get $$1+r=\frac{1-r^{2}}{1-r}\implies 1+r=1+r$$ Step 2: assuming it holds for $n=k$ then $$1+r+r^2+\ldots+r^{k}=\frac{1-r^{k+1}}{1-r}$$ Step 3: substituting $n=k+1$, we get
$$1+r+r^2+\ldots+r^{k}+r^{k+1}=\frac{1-r^{k+1+1}}{1-r}$$ $$1+r+r^2+\ldots+r^{k}=\frac{1-r\cdot r^{k+1}}{1-r}-r^{k+1}$$ $$1+r+r^2+\ldots+r^{k}=\frac{1-r\cdot r^{k+1}-(1-r)r^{k+1}}{1-r}$$ $$1+r+r^2+\ldots+r^{k}=\frac{1-r\cdot r^{k+1}-r^{k+1}+r\cdot r^{k+1}}{1-r}$$ $$1+r+r^2+\ldots+r^{k}=\frac{1-r^{k+1}}{1-r}$$ Above is true by assumption thus it holds for $n=k+1$

Hence, the given equality holds for all positive integers $n\ge 1$