Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer. 
I'm stuck at the basis step. 
If I started with $1$. I get the right hand side is $18$ which is clearly not even close. It says prove shouldn't it be always true?

Helping out with the problem.

I'm stuck at the basis step.

$p(n)$: $3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \dfrac{3(5^{n+1} -1)}{4}$. Where $n \in \{0, 1, 2, \dots  \}$.
We can rewrite the predicate. $p(n)$: $\sum_0^n3\cdot5^n = \dfrac{3(5^{n+1} -1)}{4}$
Base case: 
Here you need to start at 0 because we only  "induct" upwards.
$p(0): \sum_0^03\cdot5^n = \dfrac{3(5^{n+1} -1)}{4}$
$\implies 3 = \dfrac{3(5^1 -1)}{4} = 3 \checkmark$
Now it's up to you to show that $p(n) \implies p(n+1)$ and interpret the results.
