How do I use the principle of mathematical induction to prove whether or not $\sum_{k=1}^n (-1)^k = \frac{(-1)^n-1}2$ is a true statement?

Question

For all n elements of Natural Numbers,$\sum_{k=1}^n (-1)^k= \frac{(-1)^n-1}2$.

I proved p(1) to be true :

$\sum_{k=1}^1 (-1)^k = (-1)^1 =-1$.

And $\frac{(-1)^1-1}2 = \frac{(-2)}2 = -1$

So P(1) is true.

Where I'm stuck is proving whether or not P(n+1) is also true.

Then it becomes $\sum_{k=1}^{n+1} \frac{(-1)^{n+1}-1}2$.

How do I use Induction to prove its also true?

Answer 1

Using your notation, let $P(n)$ denote the proposition $\sum_{k=1}^{n} (-1)^k =\frac{(-1)^n-1}{2}$.

If we want to use induction to prove that $P(n)$ holds for all $n\in\mathbb{N}$ we first show $P(1)$ is true (which you've done): \begin{align} \sum_{k=1}^{1} (-1)^k =-1=\frac{(-1)^1-1}{2} \end{align}

Next we must verify that $P(n)\implies P(n+1)$. Indeed, if we assume that $P(n)$ holds, then \begin{align*} \sum_{k=1}^{n+1} (-1)^k &=\sum_{k=1}^{n}(-1)^k + (-1)^{n+1} \\ &= \frac{(-1)^n-1}{2} + (-1)^{n+1} \\ &= \frac{-(-1)^{n+1}-1}{2} + (-1)^{n+1} \\ &= \frac{(-1)^{n+1}-1}{2} \end{align*}

and thus $P(n+1)$ holds.