I am desperate for help, and far away from my institution and I had a tough time finding anything specific to my problem.
I need to prove the following:
$1 + a + a^2 +.. + a^{n-1} = \frac{a^n-1}{a - 1}$
And I am just confused how I go from here:
$\frac{a^n-1}{a-1} + a^{n+1}$ to $\frac{a^{n+1}-1}{a-1}$.
On
To prove by induction, start with the basis step (I guess you've probably done this)
$P(1): 1 = \frac{a^1 - 1}{a -1}$
which is true. For the induction hypothesis, we assume the result holds for $P(k)$
$P(k): 1 + a + a^2 + ... + a^{k-1} = \frac{a^k - 1}{a - 1}$
We want to prove that $P(k) \implies P(k+1)$, so
$1 + a + a^2 + ... + a^{k-1} + a^k = \frac{a^k - 1}{a - 1} + a^k \text{ (by induction hypothesis)}$
$\begin{align} \frac{a^k - 1}{a - 1} + \frac{a^k(a-1)}{a - 1} &= \frac{a^{k+1} +a^k - a^k - 1}{a - 1}\\ &=\frac{a^{k + 1} - 1}{a - 1} \end{align}$
which is in the required form. Hence $P(k) \implies P(k+1)$ and the result follows.
HINT: if you look carefully at the statement, it says the sum with n terms raise the n-th power of $a$. Count the number of terms in the sum carefully and you should have
$$1+a+a^2+...+a^n+a^{n+1}=\frac{a^{n+1}-1}{a-1}+\frac{(a-1)a^{n+1}}{a-1}$$