"Prove using simple induction that for each integer $n \geq 1$, $$ 5 + 5^2 + 5^3 +..... + 5^n = \frac{5^{n+1}-5}4 $$
so I start with base step
base step: $n = 1$ $$5^1 = \frac{5^{1+1}-5}4 $$ is correct.
induction step: so we assume that $P(k)$ is true for all integers $≥1$ $$ 5^{(k+1)} = \frac{5^{(k+1)+1}-5}4 $$ this is where I get stuck :X, i'm confused about the "$5 + 5^2 + 5^3 +\ldots$ " notation in question and just generally where to go from here
You have assumed that the statement is true for all $n$ uptill $k$ and now you want to prove the statement for $n = k+1$
For $n = k+1$ the summation on the LHS is
$$5 + 5^2 + \dots + 5^k + 5^{k+1}$$
We want to prove that this is equal to $\frac{5^{k+2}-5}{4}$.
Let's simplify the LHS using our induction hypothesis that the statement is true for $n = k$. We have
$$5 + 5^2 + \dots 5^k = \frac{5^{k+1}-5}{4}$$
Using the above equation below we get
$$\underbrace{5 + 5^2 + \dots + 5^k}_{\text{Use induction hypothesis}} + 5^{k+1} = \frac{5^{k+1}-5}{4} + 5^{k+1}$$ $$ = \frac{5^{k+1}-5 + 4\times5^{k+1}}{4} $$ $$ = \frac{5\times 5^{k+1}-5}{4} = \frac{5^{k+2}-5}{4}$$
This is the same as the RHS of out statement.
Q.E.D.