The Problem is under the topic 'Inclusion-Exclusion' of Combinatorics.
The 'Indicator function' $\chi_{A}(x)$ is defined as :
$\chi_{A}(x)= 1, if x \in A$ otherwise $ 0, if x \notin A$.
And has the properties :
(1) $\chi_{A \cap B}(x) = \chi_{A}(x) \chi_{B}(x)$,
(2) $\chi_{A \cup B}(x) = \chi_{A}(x) + \chi_{B}(x) - \chi_{A}(x) \chi_{B}(x)$,
(3) $\overline {\chi_{A}(x)} = 1- \chi_{A}(x)$, and analogiclly so on.
With the help of this above defined function we've to prove the following equality :
$|A_{1} \Delta \cdots \Delta A_{n}| = \sum_{i} |A_{i}| - 2 \sum_{i<j}|A_{i} \cap A_{j}| +4 \sum_{i<j<k}|A_{i} \cap A_{j} \cap A_{k}| - \cdots$
I started by using this formula :
$|A \Delta B| = |A| + |B| - 2 |A \cap B|$.
Edit (my Second attempt) :
$A_{1} \Delta A_{2} = (A_{1} - A_{2}) \cup (A_{2} - A_{1}) - (1)$
By the theorem of Inclusion- Exclusion principle, we know that :
$|A_{1} \cup \cdots \cup A_{n}| = \sum_{i} |A_{i}| - \sum_{i<j}|A_{i} \cap A_{j}| + \sum_{i<j<k}|A_{i} \cap A_{j} \cap A_{k}| - \cdots + (-1)^{n+1} |A_{1} \cap \cdots \cup A_{n}| -(2)$.
I've been trying to combine these two equations (1) and (2) to get the desired result but no success so far.
Any suggestion for this method will be very much helpful.
But, I don't know how to proceed next? Please help me!
HINT:
Use the formula $$\mathcal{X}_{A\Delta B} = \mathcal{X}_A + \mathcal{X}_B- 2\mathcal{X}_A \cdot \mathcal{X}_B$$
Then use induction on $n$ to show that $$\mathcal{X}_{A_1\Delta A_2\Delta \ldots \Delta A_n}= \sum \mathcal{X}_{A_i}- 2 \sum \mathcal{X}_{A_i} \mathcal{X}_{A_j} + 4 \sum \mathcal{X}_{A_i} \mathcal{X}_{A_j} \mathcal{X}_{A_k} -\cdots$$
$\bf{Added:}$
Note that $$1- 2\mathcal{X}_{A\Delta B} = (1-2 \mathcal{X}_A)(1-2 \mathcal{X}_B)$$
so $$1-2\mathcal{X}_{A_1\Delta A_2\Delta \ldots \Delta A_n} = \prod_{i=1}^n (1- 2 \mathcal{X}_{A_i})$$