Double induction example: $ 1 + q + q^2 + q^3 + \cdots + q^{n-1} + q^n = \frac {q^{n+1}-1}{q-1} $

Question

I'm working on a double induction problem with the following prompt:

Prove by induction on $n$ that for any real number $q > 1$ and integer $n \ge 0$: $$ 1 + q + q^2 + q^3 + \cdots + q^{n-1} + q^n = \frac {q^{n+1}-1}{q-1} $$

Based on Induction on two integer variables, I would imagine the solution is:

1. Base case $(q,n)$ as $(2,n)$ and $(q,0)$
2. Assume premise for $(2,n)$ and prove $n+1$
3. Assume premise for $(q,0)$ and prove $q+1$

But even the base cases confuse me...for example on $(q,0)$: $$1+q^0 = \frac {q^{0+1}-1}{q-1} $$ $$2 = 1 $$

That can't right. Maybe I omit the $q^0$ on the LHS, but why?

Answer 1

The problem says "for any real number $q>1$..." You can't do (standard) mathematical induction on the real numbers (see here though).

As the problem itself explicitly states, you are supposed to do induction on $n$. There is no "double induction" here. The number $q$ should be treated as just a given real number.

Answer 2

There is no double induction, as there is only one variable running through the natural numbers.

You can just leave $q$ as an indeterminate; note that the expression is $$ q^0+q^1+\dots+q^n $$ and this means that, for $n=0$, it is just $q^0=1$. So the base step is true.

For the inductive step, $$ q^0+q^1+\dots+q^n+q^{n+1}=\frac{q^{n+1}-1}{q-1}+q^{n+1}= \frac{q^{n+1}-1+q^{n+2}-q^{n+1}}{q-1} $$ and you're done.

The only restriction on $q$ is $q\ne1$ (somebody would say also for $q=0$, but if we consider $0^0=1$ the relation still holds).

Answer 3

$$\dfrac{q^{n+1}-1}{q-1}+q^{n+1}=\dfrac{q^{n+1}-1}{q-1}+\dfrac{q^{n+2}-q^{n-1}}{q-1}=\dfrac{q^{n+2}-1}{q-1}$$ and $$q^0=1=\dfrac{q^{0+1}-1}{q-1}$$ That's a proof by induction I think.