Question

Use mathematical induction to prove that if a ∈ ℝ and r≠1then a+ar^1+ar^2+...+ar^(n+1)=a(r^n+1)/(r-1).

Answer 1

Answer:

Step-by-step explanation:

In this problem we are proving the equality

$a + ar + ar^2 +\cdots + ar^n = a\frac{1-r^{n+1}}{1-r}$.

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

$n=0$: we have $a = a\frac{1-r}{1-r} = 1$. So the equality holds for $n=0$.

$n=1$: we have $a + ar =a(1+r)= a\frac{1-r^2}{1-r} = a\frac{(1-r)(1+r)}{1-r} = a(1+r)$. So the equality holds for $n=1$.

After we have checked the hypothesis for $n=0,1$ can continue to the next step.

Second: State the induction hypothesis for $n=k$. In this case the hypothesis is:

$a + ar + ar^2 +\cdots + ar^k = a\frac{1-r^{k+1}}{1-r}$.

Now, it comes the last step and, usually, the most difficult.

Third: Prove the statement for $n=k+1$ (using that the equality holds for $n=k$!). This means that we want to prove that:

$a + ar + ar^2 +\cdots + ar^k + ar^{k+1} = a\frac{1-r^{k+2}}{1-r}$.

So, let us start by the left hand side and try to get the left hand side.

$a + ar + ar^2 +\cdots + ar^k + ar^{k+1}$.

We can group the above sum in the following way

$a + ar + ar^2 +\cdots + ar^k + ar^{k+1} =(a + ar + ar^2 +\cdots + ar^k) + ar^{k+1}$.

Notice that the expression under parenthesis is the same we have in our induction hypothesis. Then,

$(a + ar + ar^2 +\cdots + ar^k) + ar^{k+1} = a\frac{1-r^{k+1}}{1-r} + ar^{k+1}$.

Now, we operate the sum that appears in the right hand side:

$a\frac{1-r^{k+1}}{1-r} + ar^{k+1} = \frac{a - ar^{k+1}+ar^{k+1}-ar^{k+2}}{1-r} = \frac{a-ar^{k+2}}{1-r} = a\frac{1-r^{k+2}}{1-r}$.

So, we have obtained that

$a + ar + ar^2 +\cdots + ar^k + ar^{k+1} = a\frac{1-r^{k+2}}{1-r}$,

which is exactly what we want to prove.