Prove by induction $ 1+3+\dots+(2n-1)=n^2 \forall n\in \Bbb N_1 $ I need help with this exercise.
What I've done so far is:
1st: Prove it with $n = 1$ (for example):

$$ (2\times1-1) = 1^2 $$
  $$ 1 = 1 $$ 

which is true
2nd: if $P(n)$ is true, then $P(n+1)$ should be true as well.
So now I substitute $n$ for $m+1$

$$ n = m+1\\
\text{and}\\
P(n) \implies P(m+1) $$
Therefore:
$$1+3+\dots+(2(m+1)-1) = (m+1)^2 \\
1+3+\dots+(2m+2-1) = (m+1)^2 \\
1+3+\dots+(2m+1) = m^2+1+2m $$

Now, the problem is that I don't know how to prove it/conclude the exercise.
Could someone help me please?
Thank you so much
If something's not very clear, please let me know. I'll try to explain again
 A: Your setup of the induction step is incorrect. You must prove that the implication $P(n) \implies P(n+1)$ is true, so it is logically incorrect to assume that this implication is true as you seem to have done. It's very important to keep track in your head at all points of the proof: which equation is known to be true, and which equation you must prove is true.
To prove $P(n) \implies P(n+1)$, you assume that $P(n)$ is true. That is, you may assume that this equation is true:
$$1 + 3 + \cdots + (2n-1) = n^2
$$
is true. Using that assumption, you must prove that this equation is true:
$$1 + 3 + \cdots + (2n-1) + (2n+1) = (n+1)^2
$$
To do that, make a substitution and do some algebra:
\begin{align*}
1 + 3 + \cdots + (2n-1) + (2n+1) &= \underbrace{1 + 3 + \cdots + (2n-1)}_{=n^2} + (2n+1)  \\
  &= n^2 + 2n + 1 \\
  &= (n+1)^2
\end{align*}
A: $$1+3+...+(2m+1) = \Big[1+3+...+(2m-1)\Big]+(2m+1)=m^2+(2m +1) =...$$
A: You've done it right, but:
$$(a+b)^2+a^2+2ab+b^2$$
So:
$$(m+1)^2=m^2+2m+1 \neq m^2+1+2$$
So, you've assumed that it's true for $n=m$, so:
$$1+3+\dots+(2m-1)=m^2$$
For $n=m+1$, you've shown it:
$$1+3+\dots+(2m+1) = m^2+1+2m $$
Now after using some more brackets and writing out more terms:
$$[1+3+\dots + (2m-1)]+(2m+1) = [m^2]+(1+2m) $$
The parts in the [] brackets are equal because of the assumption, and the only thing left to prove is that $2m+1=2m+1$, which is true.
A: You almost got it! Just a small mistake in the last step. You should note that
\begin{align*}
 1+3+...+(2m+1) = \underbrace{1+3+...+ (2m-3) + (2m-1)}_{\text{induction hypothesis!}} + (2m+1) = m^2 + (2m+1)
\end{align*}
Finally, 
$$
m^2+2m+1 = (m+1)^2
$$
so you are done.
A: Consider $n=m+1$:
$$1+3+\cdots+(2m-1)+(2(m+1)-1)=m^2+2m+2-1=(m+1)^2$$ so $P(m+1)$ holds. 
Hence by Mathematical Induction, the result is true for all $n\in\mathbb{N}$.
A: Suppose $P(n)$ is true and try to prove $P(n+1)$ to holds. 
$1+3+...+ 2n-1=n^2$ by the induction hypothesis, and now let's add the next odd number to the equation: $1+3+...+2n-1+2(n+1)-1=n^2+2n+1$
The expression on the right-hand side can be rewritten as $(n+1)^2$ and therefore the formula holds.
