Proof Problem - 6

First, let us see if this is true for $n = 1$ ,

$\frac {a(1 - q^{1 + 1})}{1 - q} = \frac {a(1 - q^2)}{1 - q} = \frac {a(1 + q)(1 - q)}{1 - q} = a(1 + q)$ ,

$a + aq = a(1 + q)$ , thus this is true for $n = 1$ ,

Now, let us assume this is true for $n = r$ so that

$a + aq + {aq}^2 + {aq}^3 + \dots + {aq}^r = a \frac {1 - q^{r + 1}}{1 - q}$ ,

Then, $a + aq + {aq}^2 + {aq}^3 + \dots + {aq}^r + {aq}^{r + 1} = a \frac {1 - q^{r + 1}}{1 - q} + {aq}^{r + 1}$ ,

That equals $a \frac {(1 - q^{r + 1}) + q^{r + 1}(1 - q)}{1 - q} = a \frac {1 - q^{r + 1} + q^{r + 1} - q^{r + 2}}{1 - q}$ ,

Which equals $a \frac {1 - q^{r + 2}}{1 - q} = a \frac {1 - q^{(r + 1) + 1)}}{1 - q}$

If this is true for $r$ and $r + 1$ , it must be true for all positive integers. Q.E.D.

It can be easily proven that $(1-q)(1+q+q^2+q^3+\dots+q^n) = 1-q^{n+1}$ by expanding them, which gives $\sum_{m=0}^n q^m-\sum_{m=0}^n q^{m+1} = 1-q^{n+1}$ Hence $a(1+q+q^2+q^3+\dots+q^n) = a(\frac{1-q^{n+1}}{1-q})$