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

Explanation:

In this problem we are proving the equality

`a + ar + ar^2 +\cdots + ar^n = a(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(1-r)/(1-r) = 1`. So the equality holds for
`n=0`.

`n=1`: we have
`a + ar =a(1+r)= a(1-r^2)/(1-r) = a((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(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(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(1-r^(k+1))/(1-r) + ar^(k+1)`.

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

`a(1-r^(k+1))/(1-r) + ar^(k+1) = (a - ar^(k+1)+ar^(k+1)-ar^(k+2))/(1-r) = (a-ar^(k+2))/(1-r) = a(1-r^(k+2))/(1-r)`.

So, we have obtained that

`a + ar + ar^2 +\cdots + ar^k + ar^(k+1) = a(1-r^(k+2))/(1-r)`,

which is exactly what we want to prove.