Union summation So for the following question: 

For $i\in\mathbf Z$ let $A_i=\{i-1,i+1\}$. Determine the following: $$\bigcup\limits_{i=1}^5 A_{2i}$$
So when we compute the union of 2i index, how would it affect the range described in Ai to get the answer?
And also, the approach which will work on the above question will also work for the following one, right?

$$\bigcup\limits_{i=1}^5 (A_{2i-1}\cap A_{2i+1})$$
Thanks in advance for all the help! :D

$A_{2i}$ = $\{2i-1,2i+1\}$, i.e. just substitute $i$ by $2i$.
Therefore for the first union 
\begin{align*}
A_2 &= \{1,3\}\\
 A_4&=\{3,5\}\\
A_6&= \{5,7\}\\
A_8&= \{7,9\}\\
A_{10}&= \{9,11\}
\end{align*}
and the union of all is $\{1,3,5,7,9,11\}$.
In the 2nd example, the union goes of the sections that always yield $2i$, therefore
the result is $\{2,4,6,8,10\}$.
