Show that $a + ar + ar^2 + \dots + ar^n = \frac{a(a-r^{n+1})}{1-r}$ for $a = 1, r = 5$
$a)~~n = 1$
$$1 + 5 = \frac{1(1-5^2)}{1-5} = \frac{-24}{-4}$$ $$6 = 6$$
$b)~~n = k+1$
$$1 + 5 + 5^2 + \dots 5^k + 5^{k+1} = \frac{1(1-5^{k+1+1})}{1-5}$$ $$\frac{1(1-5^{k+1})}{1-5} + 5^{k+1} = \frac{1(1-5^{k+1+1})}{1-5}$$
Here is where it starts to fall apart when I am trying to simplify both sides. I think that the best approach is to take the logs of both sides, but my logarithmic skills are subpar at best. Here is what I tried to do, but I don't think I am using the properties correctly.
$$log\bigg(\frac{1(1-5^{k+1})}{1-5}\bigg) + log(5^{k+1}) = log\bigg(\frac{1(1-5^{k+1+1})}{1-5}\bigg)$$ $$log(1) - (k+1)log(5) - log(-4) + (k+1)log(5) = log(1) - (k+1+1)log(5) - log(-4) $$
I think that I am maybe not splitting the numerators and denominators up properly?
Without using $\log$ we have $$ \frac{1-5^{k+1}}{1-5}+5^{k+1}=\frac{1-5^{k+1}+(1-5)5^{k+1}}{1-5}=\frac{1-5^{k+1}+5^{k+1}-5^{k+2}}{1-5}=\frac{1-5^{k+1+1}}{1-5} $$ as claimed.
Watch out when using the $\log$, in particular recall that $\log(a+b)\neq\log(a)+\log(b)$.