Prove by PMI that:
$(-1)^{1}+(-1)^{2}+(-1)^{3}+⋯+(-1)^{n}=\frac{(-1)^{n}-1}{2}$
Induction Hipotesis for some integer k:
$(-1)^{1}+(-1)^{2}+(-1)^{3}+⋯+(-1)^{k}=\frac{(-1)^{k}-1}{2}$
Induction Step:
$(-1)^{1}+(-1)^{2}+(-1)^{3}+⋯+(-1)^{k}+(-1)^{k+1}=\frac{(-1)^{k+1}-1}{2}$
So we have:
$\frac{(-1)^{k}-1}{2}+(-1)^{k+1}=\frac{(-1)^{k+1}-1}{2}$
Working with LHS:
$\frac{(-1)^{k}-1}{2}+(-1)^{k+1}=\frac{(-1)^{k}-1+2(-1)^{k+1}}{2}$
$=\frac{(-1)^{k}-1+2(-1)^{k+1}}{2}$
$=\frac{(-1)^{k}[2(-1)^{1}+1]-1}{2}$
I'm not sure how to proceed since $2(-1)^{1}+1=-1$, so LHS will end up being $=\frac{-(-1)^{k}-1}{2}$
Thanks in advance.
Hi I would first of all rewrite the task to
$$ \sum_{i=1}^n (-1)^{i}=\frac{(-1)^n-1}{2}$$
So at first for n=1 the statement is correct since
$$(-1)^1=\frac{(-1)^1-1}{2}$$
Then the induction step is then for all $n\geq 1$
$$\sum_{i=1}^{n+1} (-1)^k=\sum_{i=1}^{n} (-1)^k+(-1)^{n+1}$$ $$=\frac{(-1)^n-1}{2}+(-1)^{n+1}=\frac{(-1)^n-1}{2}+\frac{2(-1)^{n+1}}{2}$$ $$\frac{(-1)^{n+1}(\frac{1}{-1}+2)-1}{2}=\frac{(-1)^{n+1}-1}{2}$$