How can i proof by Mathematical Induction? $\sum_{i=1}^{n} (2i-1) = n²-1$ to n>=2 $\sum_{i=1}^{n} (2i-1) = n²-1$ to n>=2
Proof:
first step:$\sum_{i=2}^{n=2} (2i-1) = (2*2)-1 = 3$ and $n²-1 = (2*2)-1=3$
so $\sum_{i=2}^{n=2} (2i-1) = n²-1$
$\sum_{i=1}^{k+1} (2i-1) = \sum_{i=1}^{k} (2i-1)+ 2(k+1)-1$
$=k²-1+2k+1$
$=k²+2k$
$=(k+1)²-1$
That its correctly?
 A: We have
$\sum_{i=1}^n (2i-1)=\left(2\sum_{i=1}^n i\right) -n=n(n+1)-n=n^2$.
So what you are trying to prove is impossible.
There are some flaws in your induction.
For example the beginning of the induction should be $n=2$, and then the sum you have to evaluate is
$\sum_{i=1}^2 (2i-1)=(2-1)+(4-1)=4$.
Your induction is over $n$, and the assumption already fails here, as you can see.
A: It isn't true.  And I think deep down you knew that.
I think you probably tried to do a base case $\sum_{i=1}^{n=1}(2i-1) = 1^2-1$ and got on the LHS $\sum_{i=1}^{n=1}(2i-1) = 2\cdot 1-1 =1$ and got on the left hand side and got $1^2-1 = 0$.  I think you probably figured maybe it was one of those cases where the base case of $0$ or $1$ is a discrepancy so your probably figured you try it with a base case of $n=2$.
[EDIT: Oh, I see the premise was $n \ge 2$ so you probably started with $n=2$ as in the next paragraph.]
So you tried $\sum_{i=1}^{n=2}(2i-1) = 1+3 = 4$ but $4 \ne 2^2 -1$.  But I think you probably noticed that $3$ does equal $2^2-1$ and you figure if you didn't have that $1$ it would work and if you started at $i=2$ rather than $i=1$ you wouldn't have the $1$ and everything would work.
So you did as a base case $\sum_{i=2}^{n=2}(2i-1) = 3 = 2^2-1$ and declared it a success.
But here's the problem.  That's a fudge and fudges will catch up with you.  If your base case is $\sum_{i=2}^{n=2}(2i-1)=2^2-1$.  Then in your induction step you can NOT assume $\sum_{i=1}^{n}(2i-1)=n^2-1$ because that WASN'T what your base case was.
The induction step MUST match the base case and so your induction step you will have proven that if $\sum_{i=2}^n (2i-1) = n^2 -1$ then $\sum_{i=2}^{n+1}(2i-1) =(n+1)^2 -1$.
And you have successfully proven that $\sum_{i=2}^n(2i-1) = n^2 -1$.
But you HAVEN'T proven $\sum_{i=1}^2(2i-1) = n^2-1$.  You haven't proven that because IT ISN"T TRUE.
If we try $n=3$ we get $1 + 3 + 5 = 9 \ne 3^2 -1$ and for $n=4$ we get $1+3+5+7=16\ne n^2-1$ and so on.  The statement you are trying to prove is NEVER true.
The lesson to be learned here is that is you find something doesn't work-- don't fudge.  Try to see if maybe the problem is wrong. (In this case it was wrong.) Admittedly, in most cases it probably won't be wrong and we will be... but even so, we can never fix it by fudging.  Fudging always catches up with us.
.....
SO, anywho...... You have actually proven that $\sum_{i=2}^n (2i -1) = n^2 -1$.  That means that $\sum_{i=1}^n(2i-1) = (2\cdot 1-1)+\sum_{i=1}^n(2i-1) = 1 + (n^2 -1) = n^2$.
And that's the actual result.  $\sum_{i=1}^n (2i-1) = n^2$.
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EDIT:  I realize later that, I think you misread the problem.  I don't think that is a minus sign.  I think it is suppose to be a dash.  I think what you are supposed to prove is:  Prove by mathematical induction that $\sum_{i=1}^n (2i-1) = n^2$ --- for $1$ to $n \ge 2$.
