Answer:
Step-by-step explanation:
In this problem we are proving the equality
[tex] a + ar + ar^2 +\cdots + ar^n = a\frac{1-r^{n+1}}{1-r}[/tex].
When we want prove an equality by induction we need to follow some steps.
First: Check the hypothesis for some initial cases. In this exercise we take
[tex]n=0[/tex]: we have [tex]a = a\frac{1-r}{1-r} = 1[/tex]. So the equality holds for [tex]n=0[/tex].
[tex]n=1[/tex]: we have [tex]a + ar =a(1+r)= a\frac{1-r^2}{1-r} = a\frac{(1-r)(1+r)}{1-r} = a(1+r)[/tex]. So the equality holds for [tex]n=1[/tex].
After we have checked the hypothesis for [tex]n=0,1[/tex] can continue to the next step.
Second: State the induction hypothesis for [tex]n=k[/tex]. In this case the hypothesis is:
[tex] a + ar + ar^2 +\cdots + ar^k = a\frac{1-r^{k+1}}{1-r}[/tex].
Now, it comes the last step and, usually, the most difficult.
Third: Prove the statement for [tex]n=k+1[/tex] (using that the equality holds for [tex]n=k[/tex]!). This means that we want to prove that:
[tex]a + ar + ar^2 +\cdots + ar^k + ar^{k+1} = a\frac{1-r^{k+2}}{1-r}[/tex].
So, let us start by the left hand side and try to get the left hand side.
[tex] a + ar + ar^2 +\cdots + ar^k + ar^{k+1}[/tex].
We can group the above sum in the following way
[tex] a + ar + ar^2 +\cdots + ar^k + ar^{k+1} =(a + ar + ar^2 +\cdots + ar^k) + ar^{k+1} [/tex].
Notice that the expression under parenthesis is the same we have in our induction hypothesis. Then,
[tex](a + ar + ar^2 +\cdots + ar^k) + ar^{k+1} = a\frac{1-r^{k+1}}{1-r} + ar^{k+1}[/tex].
Now, we operate the sum that appears in the right hand side:
[tex]a\frac{1-r^{k+1}}{1-r} + ar^{k+1} = \frac{a - ar^{k+1}+ar^{k+1}-ar^{k+2}}{1-r} = \frac{a-ar^{k+2}}{1-r} = a\frac{1-r^{k+2}}{1-r}[/tex].
So, we have obtained that
[tex]a + ar + ar^2 +\cdots + ar^k + ar^{k+1} = a\frac{1-r^{k+2}}{1-r}[/tex],
which is exactly what we want to prove.
