Prove by Induction (Geometric Progression): $\sum_{i=0}^n q^i=\frac{q^{n+1}-1}{q-1}$

Question

Prove by induction that for any real number $q≠1$ and any $n\in \mathbb N$ we have $ \sum_{i=0}^n q^i=\frac{q^{n+1}-1}{q-1} $

Answer 1

Base Case:

For $n=0$, our sum is $\sum_{i=0}^n q^i$ is equal to 1, which is also the value of $\frac{q^{n+1}-1}{q-1}$ at $n=0$. Hence base case justified.

Induction hypothesis:

Let

$$\sum_{i=0}^k q^i=\frac{q^{k+1}-1}{q-1}$$ for some $k \ge0$.

Inductive step:

$$\sum_{i=0}^{k+1} q^i=\frac{q^{k+1}-1}{q-1} + q^{k+1}$$

Taking it to a common denominator, and some simple algebraic manipulation.

$$\sum_{i=0}^{k+1} q^i=\frac{q^{k+2}-1}{q-1}$$

Q.E.D