I'm trying to learn how to prove by induction Please, I need some help to prove this
$$1+\sum_{i=1}^n 5^i = \frac{5^{n+1}-1}{4}$$
I got this but I'm not sure how to proceed
Base n=1
$$=\frac{(5^2)-1}{4}$$
$$=\frac{25-1}{4}$$
$$=\frac{24}{4}$$
$$={6}$$
It works for n=1
Induction hypothesis
Let n=k
$$1+5+25+125+...+k=\frac{5^{k+1}-1}{4}$$
Induction step
Let n=k+1
Prove $$1+5+25+125+...+k+(k+1)=\frac{5^{k+2}-1}{4}$$
By induction hypothesis
$$1+5+25+125+...+k=\frac{5^{k+1}-1}{4}$$
And I tried this way using laws of exponents
$$1+5+25+125+...+k+(k+1)=\frac{5^{k+1}-1}{4}+(k+1)$$
$$=\frac{5^{k+1}-1}{4}+\frac{4(k+1)}{4}$$
$$=\frac{5^k*5+4k+3}{4}$$
I think that I'm doing it wrong but  would be so grateful if you could help me
EDIT: I just realized how bad my summation was, but now I understand the proof, ty all
 A: It seems to me, you interpret the summation is wrong.
But, you have correctly confirmed that the formula is correct for $n=1.$
You can continue from where you left off.
For $n=k$, you get
$$1+5+25+\cdots+5^k=\frac{5^{k+1}-1}{4}$$
Then for $n=k+1$, you get
$$\begin{align}1+5+25+\cdots+5^k+5^{k+1}
&=\frac{5^{k+1}-1}{4}+5^{k+1}\\
&=\frac{4\times 5^{k+1}+5^{k+1}-1}{4}\\
&=\frac{5\times 5^{k+1}-1}{4}\\
&=\frac{5^{k+2}-1}{4}\\
&=\frac{5^{(k+1)+1}-1}{4}.\end{align}$$
This means, if the statement is correct for  $n=k$, then it is also correct for $n=k+1$.
Then by induction, you are done.
A: Let $k\in\Bbb N$ and assume that
$$
1+\sum_{i=1}^k5^i=\frac{5^{k+1}-1}4.
$$
Then
\begin{align}
1+\sum_{i=1}^{k+1}5^i&=\left(1+\sum_{i=1}^k5^i\right)+5^{k+1}\\
&=\frac{5^{k+1}-1}4+5^{k+1}\\
&=\frac{5^{k+1}-1+4\times5^{k+1}}4\\
&=\frac{5\times5^{k+1}-1}4\\&=\frac{5^{k+2}-1}4.
\end{align}
A: Other answers are focused on the inductive step (which is almost always the harder part). But I'm going to assert that your basis step also needs correction. You write (shown in one line for brevity):
$$=\frac{(5^2)-1}{4} =\frac{25-1}{4} =\frac{24}{4} ={6}$$
The following observations are things I commonly need to point out to my students:

*

*What's on the left-hand side of that initial equal sign? You cannot start a statement with a relation symbol. You wouldn't write a statement "$= 7$" ("is equal to 7"); that's a sentence fragment, doesn't communicate any information, and needs to be fixed.

*Related: The formula in the theorem has two sides! You've shown (once the preceding gets fixed) that when $n = 1$, the right-hand side is equal to $6$. But is that the same as the left-hand side? Are they actually equal? That remains to be verified.

Personally, when proving summation formulas, I find it clearest to write chained relations for the left-hand and right-hand sides separately, then confirm they are the same value. (And of course on the blackboard many abbreviate these as LHS and RHS.) In this case you should be able to write something like:
When $n = 1$:

*

*The left-hand side of the formula is $1+\sum_{i=1}^1 5^i = ... = 6$.

*The right-hand side of the formula is $\frac{(5^{1+1})-1}{4} = ... = 6$.

*So the equality is true when $n = 1$.

