By mathematical induction I need to show that $a$ is a positive real number other than $1$, then
$$\sum^n_{j=1}{a^j}=(a)\frac{1-a^n}{1-a}$$
For each natural number $n$.
We us ethe first principle of mathematical induction. For the base case that $1^3=1=(\frac{1*2}{2})^2$. We presume that the above is true for all positive real numbers other than 1. Then
$$\sum^{n+1}_{j=1}{a^j}=(a)\frac{1-a^{n+1}}{1-a}$$
Could I show this and then somehow expand and factor? Or am I on the wrong track?
Your base case is incorrect: it should be for $n=1$ (and $a\neq 1$): $$\sum_{j=1}^1 a^j = a = (a)\frac{1-a^1}{1-a}.$$
Then, assume true for $n>1$: $$\sum_{j=1}^n a^j = (a)\frac{1-a^n}{1-a},$$ For $n+1$: $$\sum_{j=1}^{n+1} a^j = \sum_{j=1}^n a^j + a^{n+1},$$ by definition, so plug in the expression for $n$: $$\sum_{j=1}^{n+1} a^j = (a)\frac{1-a^n}{1-a} + (a) a^n,$$ and simplify the RHS of this expression to prove the result.
You've misunderstood the question. You aren't trying to prove it is true for all real $a$; you are trying to prove the statement for all natural $n$.
You set $a > 0$, $a \ne 1$ arbitrarily once and keep it fixed and prove that for all $n$
$\sum_{j=1}^n a^j = a\frac{1 - a^{n}}{1-a}$.
Step 1: Show for $n = 1$.
$\sum_{j=1}^1 a^j = a^1 = a = a*1 = a\frac{1-a}{1-a}$. (This is why $a \ne 1$.)
Step 2: Assume true for $n =k$.
Show it is true for $n = k+1$.
Hint: $\sum_{j=1}^{k+1} a^j = a^{k+1} + \sum{j=1}^{k} a^j = a^{k} + a\frac{1-a^k}{1-a}$.
So you need to prove $a^{k} + a\frac{1-a^k}{1-a} = a\frac{1 - a^{k+1}}{1- a}$.
Can you do that.
