Note $p_{k}=\sum_{1 \leq i_{1} < i_{2} \dots < i_{k} \leq i_{n} } P(A_ {i_{1}} \cap A_ {i_{2}} \cap \dots \cap A_ {i_{k}})$ prove $P(\cup^n_{k=1}A_k) = \sum^n_{k=1}(-1)^{k-1}p_k$
This question uses a result from a previous question, it says that,
by developing the last term in the result (2) :
$ \mathbb{1}_{ \cup^n_{k=1} A_k } = 1 - \prod^n_{k=1}( 1 -\mathbb{1}_{A_k} ) \dots (2) $
we get :
$ \prod^n_{k=1}( 1 - \mathbb{1}_{A_k} ) = 1 - \sum^n_{k=1} (-1)^{k-1}\sum_{1 \leq i_{1} < i_{1} \dots < i_{k} \leq i_{n} }$ $ \mathbb{1}_{ A_{ i_{1} } \cap A_{i_{2}} \cap \dots \cap A_{i_{k} } }$
using linearity of expactation we get: $P(\cup^n_{k=1}A_k) = \sum^n_{k=1}(-1)^{k-1}p_k$
Remark included in the solution at the end : you also need to notice that $E( \mathbb{1}_{A} ) = P(A)$
Now my problem in order to understand this proof, I need to understand how this double summation works because what I come to is, by the inclusion exclusion theorem :
$P(\cup^n_{k=1}A_k) = \sum^n_{k=1}P(A_k) + (-1)^{2+1} P(A_{i_{1}}\cap A_{i_{2}}) + (-1)^{3+1} P(A_{i_{1}} \cap A_{i_{2}} \cap A_{i_{3}} )+ \dots + (-1)^{n+1}P(A_{i_{1}} \cap \dots \cap A_{i_{n}} ) $
How does this summation work so that it is equivalent to : $P(\cup^n_{k=1}A_k) = \sum^n_{k=1}(-1)^{k-1}p_k$
How does it gets out the $\sum (-1)^{k-1}$ to sum then over the $p_{k}=\sum_{1 \leq i_{1} < i_{2} \dots < i_{k} \leq i_{n} } P(A_ {i_{1}} \cap A_ {i_{2}} \cap \dots \cap A_ {i_{k}})$ and why k-1 ?
I first thought of $\sum^n_{k=1}P(A_k) = 1$
I spent enough hours trying to get myself an illustration of what is happening using the definitions of double sums but I dont get the right result