How to do induction proof for all $n \ge 1$, $((-1)^n+1)n = (1-(-1)^n(2n+1))$?

Question

Having trouble with this proof by induction:

$\forall n \in \mathbb{Z}: n \ge 1, $

$$1 - 2 + 3 - 4 +... + (-1)^{n+1}n = \frac{1-(-1)^{n}(2n+1)}{4} $$

So far I've got to this stage below but I've had trouble simplifying the left side to match the right:

$$\frac{1-(-1)^k (2k+1)}{4} + (-1)^{k+2}(k+1) = \frac{1-(-1)^{k+1}(2(k+1)+1)}{4} $$

$$\frac{1-(-1)^k (2k+1) + 4((-1)^{k+2}(k+1))}{4} = \frac{1-(-1)^{k+1}(2k+3)}{4} $$

Does anyone have any solutions?

Answer 1

Just note that \begin{align*} 1 - ( - 1)^k (2k + 1) + 4( - 1)^{k + 2} (k + 1) &= 1 - ( - 1)^{k + 1} ( - (2k + 1) + 4(k + 1)) \\ & = 1 - ( - 1)^{k + 1} ( - 2k-1 + 4k+4) \\& = 1 - ( - 1)^{k + 1} (2k + 3)\\& = 1 - ( - 1)^{k + 1} (2(k+1) + 1). \end{align*}