Find $\sum{(-1)^22n }$ from $n=0$ to $n=28$

Question

Find $\sum{(-1)^22n }$ from $n=0$ to $n=28$

I can't find a formula for alternation series

Answer 1

It is

$$\sum{(-1)^22n }=\sum{2n }=2\sum{n }=2\times \frac{(1+28)\times 28}{2} =28\times 29$$

Answer 2

Hint: Assuming your mean $\sum_{n=0}^{28}(-1)^n2n$, note that you can combine each even term with the odd one after it and get $-2$. How many pairs are there? Then you didn't have anything to combine $n=28$ with, so add that in by hand.

Answer 3

If it's an alternating series I assume you mean $$S=\sum_{n=0}^{28}(-1)^n(2n)\ .$$ If this is the case, hint: group the terms in pairs as far as possible, $$S=(0-2)+(4-6)+(8-10)+\cdots+(52-54)+56\ .$$ Can you take it from here?