Prove by induction for the sum of $\sum_{k=1}^{2n-1} (-1)^{k-1}(k-1)(k+1) = 2n^2 - n - 1$ I'm trying to prove following equation with induction:
$$
\sum_{k=1}^{2n-1}  (-1)^{k-1}(k-1)(k+1) = 2n^2 - n - 1
\\~~\\
$$
First step is the base case, which for $ n = 1$ is true.
$$
(-1)^{0}(0)(2) = 0 = 2(1)^2-1-1
$$
Then we can assume that the equation is true for $n$, therefore we want to prove that it's also the case for $n+1$
$$
\sum_{k=1}^{2(n+1)-1}  (-1)^{k-1}(k-1)(k+1) = \\ (\sum_{k=1}^{2n-1} (-1)^{k-1}(k-1)(k+1) ) + (-1)^{2(n+1)-1}(2(n+1)-2)(2(n+1))
$$
That means for the right side of the equation:
$$
2n^2-n-1 + (-1)^{2(n+1)-1}(2(n+1)-2)(2(n+1))
$$
Removing the brackets results in:
$$
2n^2-n-1+(-1)^{2n}(2n)(2n+2) = \\~\\2n^2-n-1+(-1)^{2n}(4n^2+4n)
$$
I'm not sure if I am allowed to make an assumption, but I did: 
$$
\forall n \epsilon \mathbb{N} :(-1)^{2n} = 1
$$
Since it's equivalent to:
$$
((-1)^n)^2 =  ((-1)^2)^n
$$
By replacing it with $1$, the last step is summing it up:
$$
6n^2+3n-1
$$
Which is wrong ... Am I allowed to make the assumption? And where did I make the mistake? (Hint preferred) 
 A: When you substitute $n+1$ into the upper limit of summation (above the $\sum$ symbol), you get there $2(n+1)-1$. But you should simplify this expression: $2(n+1)-1=2n+1$.
Now, this sum ranges from $k=1$ up to $k=2n+1$. You did exactly the right thing by extracting the sum that goes from $k=1$ up to $k=2n-1$. But how many more terms are there, after $k=2n-1$? Hint: more that one.
A: First of all, correct notations is 
$$\sum_{k=1}^{2n-1}(-1)^{k-1}(k-1)(k+1) = 2n^2 - n - 1$$
Then, what you did in the base case is correct. After that, if we suppose $n \ge 2$ and argument holds for all $n$, for $n+1$ we have 
$$\sum_{k=1}^{2(n+1)-1}(-1)^{k-1}(k-1)(k+1)$$
Notice that we can write this sum as $\sum_{k=1}^{2n-1}(-1)^{k-1}(k-1)(k+1) + $ [the expression for $k = 2n$] $+$[the expression for $k = 2n+1$], which yields:
$$= \sum_{k=1}^{2n-1}(-1)^{k-1}(k-1)(k+1)+(-1)^{2n-1}(2n-1)(2n+1)+(-1)^{2n}(2n)(2n+2)$$
$$= 2n^2-n-1-(4n^2-1)+(4n^2+4n)$$
$$= 2n^2+3n$$
$$ = 2(n+1)^2-(n+1)-1$$
A: $$[(k+1)^2-1]-[k^2-1]=2k+1$$
For the rest you can just telescope.
