Hey i'm stuck at a question of summations which i don't understand how to prove

Question

The questions is to prove the following summation for every positive integer n: $$\sum_{i=1}^{2n}(1+(-1)^i)=2n$$ https://i.ibb.co/dt77GGK/dx.png

Answer 1

Hint:

$$\sum_{i=1}^{2n}{(1+(-1)^i)}=\sum_{i=1,i {\text{ odd}}}^{2n}{((-1)^i+1)}+\sum_{i=1,i {\text{ even}}}^{2n}{((-1)^i+1)}$$ What is $(-1)^i+1$ when $i$ is odd?

What is $(-1)^i+1$ when $i$ is even?

How many even numbers are there between $1$ and $2n$?

Answer 2

The $(-1)^i$'s alternate and cancel out in pairs. You just get $\sum_{i=1}^{2n}1=2n$.

Answer 3

Looks to me like a candidate for "proof by induction"! When n= 1 this is $\sum_{i=1}^2 (1+(-1)^i)= (1+ (-1)^1)+ (1+ (-1)^2)= (1- 1)+ (1+ 1)= 2= 2(1)$.

Now assume that, for some k, $\sum_{i=1}^{2k} (1+ (-1)^i)= 2k. $\sum_{i= 1}^{2(k+ 1)} (1+ (-1)^i)= \sum_{i=1}^{2k+2}\sum_{i=1}^k (1+ (-1)^i)+ (1+ (-1)^{k+1})+ (1+ (-1)^{k+2})= 2k+ (1+ (-1)^{k+1})+ (1+ (-1)^{k+2}$.

Complete this as two cases: k even or k odd. If k is even then k+ 1 is odd and k+2 is even. The sum is 2k+ (0)+ (2)= 2k+ 2= 2(k+1).

If k is odd then k+ 1 is even and k+ 2 is odd. The sum is 2k+ (2)+ (0)= 2k+ 2= 2(k+ 1).