Prove by induction. $n \in \mathbb N$.

Question

$$\sum_{i=0}^{n-1} ar^i = \frac{a(r^n-1)}{r-1}$$ for $r \in \mathbb R, r \ne 1, n \ge 1 $

i. Show true for $n=1$ $$\frac{a(r^1-1)}{r-1} = a$$

ii. I am stuck on how to do the second part of the proof, and the logic behind it.

Answer 1

It is quite common, when you try to do induction with sums, that you separate the first part of the sum and the last term, in order to apply the induction hypothesis. In this case, assuming that the formula holds for n-1, you can continue the following computation: $$\sum_{i=0}^n ar^i = \sum_{i=0}^{n-1}ar^i + ar^n = \frac{a(r^n-1)}{r-1} + ar^n=\dots$$

Answer 2

Hint: Like in my comment the $a$ coefficient is not necessary. Assume the equation is true for all $n\leq N$. Then do the induction step:

$N \to N+1$:

$$\sum_{i=0}^{(N+1)-1}r^i= \sum_{i=0}^{N-1}r^i+r^N.$$ Now use the formula for the sumtion with the upper bound $N-1$.

$$\sum_{i=0}^{(N+1)-1}r^i= \frac{r^N-1}{r-1}+r^N=\frac{r^N-1+r^N(r-1)}{r-1}=\frac{r^{N+1}-1}{r-1}$$

The last line completes the induction.

Answer 3

Base case: $n=1$, $$\frac{a(r^1-1)}{r-1} = a$$

Assume true $$\sum_{i=0}^{n-1} ar^i = \frac{a(r^n-1)}{r-1}$$ for $r \in \mathbb R, r \ne 1$, up to some $n > 1.$

Then, $$\sum_{i=0}^{n} ar^i = ar^n+\sum_{i=0}^{n-1} ar^i=ar^n+\frac{a(r^n-1)}{r-1}=\frac{ar^n(r-1)+a(r^n-1)}{r-1}=\frac{a(r^{n+1}-1)}{r-1},$$ and we're done.