Number of leaves in a tree (all types of trees) I have to prove the following claim, given the tree $T=(V,E)$, $|V|=n\geq2$:
 $$|V_1| = 2 + \sum (j-2)|V_j|$$
where the sum is from $j=3$ up to the highest degree, and
$$V_i = \{ x \in V \mid \deg(x) = i\}.$$
This was a bonus question given by my professor. We were sitting on this question for hours and have no idea how to prove it.
Can someone help out with a hint?
Thanks!
 A: Prove it by induction on $n$:


*

*It's trivial for the case $n=2$. (Why?)

*The inductive step is to add a vertex and, because it's a tree, you've changed at most three of the $V_i$. (Which ones? How have they changed? Understand this before proceeding.) In the sum, then, you're adding $1$ to the left-hand side and $-(k-2)+(k+1-2)$ for some $k$ to the right, which maintains the equality.

A: Count the number of edges in two ways.


* By the Handshake Lemma: $|E|=\frac{1}{2}\sum\limits_{i=1}^\infty i|V_i|$.

* By the characterization of trees: $|E|= \sum\limits_{i=1}^\infty |V_i|-1$.
Equate these two quantities, and remember that $|V_1|$ is the number of leaves.
$\frac{1}{2}\sum\limits_{i=1}^\infty i|V_i|=  \sum\limits_{i=1}^\infty |V_i|-1,$
so
$\sum\limits_{i=1}^\infty (i-2)|V_i|+2=0$
so
$-|V_1|+0+\sum\limits_{i=3}^\infty (i-2)|V_i|+2=0$
so
$|V_1|=\sum\limits_{i=3}^\infty (i-2)|V_i|+2$
which is what you are after.
A: There is a close relationship between your problem and the degree sum formula for graphs.  For any finite graph, tree or otherwise, the equation
$$2|E| = \sum_{v\in V} \deg(v)$$
always holds.  If you haven't proved this before, it's worth stopping to think about why the equation is correct.
It is possible to transform the equation
$$|V_1| = 2 + \sum (j-2)|V_j|$$
into an explicit statement of the degree sum theorem for trees.  Hint: $j|V_j|$ is just another way to write $\sum_{v\in V_j} \deg(v)$.
