(-1)+(-1)^2+(-1)^3+...+(-1)^n= ((-1^n)-1)/2

Prove that the statement is true for every positive integer n.

Let n = 1
(-1)^1 = -1
(-1^1 - 1)/2 = -2/2 = 1
Therefore, the hypothesis is true for n = 1.
Suppose the hypothesis is true for n and consider n + 1.
-1 + (-1)^2 + ... + (-1)^n + (-1)^(n + 1) =
(-1^n - 1)/2 + (-1)^(n + 1)
If n is even, (-1^n - 1)/2 + (-1)^(n + 1) =
(1 - 1)/2 - 1 = (1 - 1)/2 - 2/2 = (1 - 1 - 2)/2 = (-1 - 1)/2 =
(-1^(n + 1) - 1)/2
If n is odd, (-1^n - 1)/2 + (-1)^(n + 1) =
(-1^n - 1)/2 + 1 = (-1 - 1)/2 + 1 = (-1 - 1)/2 + 2/2 = (-1 - 1 + 2)/2 =
(1 - 1)/2 = (-1^(n + 1) - 1)/2
Therefore, the hypothesis being true for n implies it is true for n + 1.
Therefore, the hypothesis is proven by induction.

(-1)+(-1)^2+(-1)^3+(-1)^n= ((-1^n)-1)/2
-1+1-1+(-1)^n = ((-1^n)-1)/2
-1+(-1)^n = ((-1^n)-1)/2

If n is even, -1^n will equal 1.

-1 + 1 = (1-1)/2
0 = 0/2
0 = 0

If n is odd, -1^n will equal -1.

-1 - 1 = (-1 - 1)/2
-1 - 1 = -2/2
-2 = -1

Are you leaving something out? It works for positive even numbers, but not for positive odd numbers.

induction my son