Why does the formula 2n + 1 find the child node in a binary heap?
I would like to propose my revisited version of Hiroki's answer.
Currently it's been sitting in peer-review (https://cs.stackexchange.com/review/suggested-edits/66932) for a while, so it's not being of use to anyone.
Credits to Hiroki for the original answer and structure.
I've simply clarified the math and made every step explicit and chosen a slightly different terminology.

Let's assume that each tier of the heap is an array.
T1 [                n0                 ]

T2 [      n1,                n2        ]

T3 [  n3,     n4,       n5,       n6   ]

T4 [n7, n8, n9, n10, n11, n12, n13, n14]

It may not be clear, but I want you to imagine that $n0$ is the parent of $n1$ and $n2$.
Likewise,  $n1$ is the parent of $n3$ and $n4$.
For example, $T2$, the second tier, contains nodes $[n1, n2]$ and has length $2$.
Let's say $i$ is the global index of the node in question (i.e. the node's index if the entire tree were to be collapsed within a single array), and
$j$ is the local index of the node, i.e. the index of the node within its tier.
For example, in the diagram above, if the node in question is $n3$, $i = 3$ and $j = 0$, because $n3$ is the first element in the 3rd tier.
i = global index of a node n
j = local index of a node n within the tier where the node exists
T = tier

The maximum number of nodes in a tree up to a certain tier can be expressed by:
$total\_nodes\_up\_to(T) = 2^T - 1$
e.g. when you have 3 tiers, you can have at most 7 nodes, as $2 \cdot 2 \cdot 2 - 1 = 7$
This is because there are $2^{T-1}$ nodes in each tier (note: tier numbering in our example arbitrarily starts from $1$, not $0$) and the sum of powers of $2$ up to $n$ is equal to $2^{n+1} - 1$, as you can see here https://math.stackexchange.com/questions/1990137/the-idea-behind-the-sum-of-powers-of-2
Since indexes start from 0, the index of the last node in an array of nodes will be $numNodes - 1$. This means that the global index of the last node in some tier $T$ is:
$i_{last} = total\_nodes\_up\_to(T) - 1 = 2^T-1-1$
While the local index of the last node in the tier is:
$j_{last} = 2^{T-1} - 1$ // since there are $2^{T-1}$ elements in tier $T$
We can now compute the global index of the first node in the tier by subtracting the local index of the last node in the tier from its global index. You can visualize it mentally as &quot;walking backwards&quot; starting from the the last node - if you take local-index-many steps back you land onto the first node in the tier:
$i_{first} = i_{last} - j_{last}$
$i_{first} = 2^T - 1 - 1 - j_{last}$
$i_{first} = 2^T - 1 - 1 - (2^{T-1} - 1)$
$i_{first} = 2^T - 2^{T-1} - 1$
$i_{first} = 2*2^{T-1} -2^{T-1} - 1$
$i_{first} = (2 - 1)*2^{T-1} - 1$
$i_{first} = 2^{T-1} - 1$
We can now compute the global index of a node $n$ by adding its local index and the global index of the first node in its tier:
$i = i_{first} + j$
$i = 2^{T-1} - 1 + j$
Let's now think about the first (left) child of the node in question.
The left child, n', will be in the next tier, T+1. 
The indices in T+1 will be referred to as i', j', i'_first ...

Based on what we've shown so far, we can say that the global index of the first node in the next tier is:
$i_{first}' = 2^{T+1-1} - 1 = 2^T - 1$
Now, for each predecessor of the parent node $n$ in tier $T$ there will be 2 predecessors of left-child node $n'$ in $T+1$. Also, given index $j$ in $T$, we know there are $j$ predecessors of $n$ in $T$ - i.e. the index of a node is equal to the number of its predecessors. So we can conclude that:
$j' = 2j$
Putting it all together, we can conclude that
$i' = i_f' + j'$
$i' = 2^T - 1 + 2j$
Now let's rework the previous equation for the global index of parent node $n$, $i = 2^{T-1}-1+j$ to the following:
$i + 1 = 2^{T-1} + j$
Finally, let's compare that to the equation of the global index of the left-child node:
$\begin{align}
i' &amp;= 2^T - 1 + 2j\\ 
   &amp;= 2*2^{T-1} + 2*j - 1\\
   &amp;= 2*(2^{T-1} + j) - 1\\
   &amp;= 2*(i + 1) - 1      \text{//NOTE: Here we use $i + 1 = 2^{T-1} + j$ mentioned above}\\
   &amp;= 2i + 2 - 1\\
   &amp;= 2i + 1\\
\end{align}$
Although the question was why the $2n + 1$ formula works, note that I used $i$ instead of $n$!
Let me know if there is anything unclear.