I'm having trouble with a proof. I can't see where to go next. Or if I'm using the wrong variable (maybe I should reason from the point of deg(vi) instead of n?).
The question is:
"Let $T$ be a tree of order $n>1$. Show that the number of leaves is
$2+\sum_{deg(v_i)\geqslant3}(deg(v_i)-2)$
where the sum is over all vertices of degree 3 or more."
I've started with proof by induction. With step 1:
Suppose $n=2$, Then, since neither vertex has degree 3 or more, the sum equals 0. So
$2+\sum_{deg(v_i)\geqslant3}(deg(v_i)-2)=2+0=2$ There are indeed 2 leaves in a tree with $n=2$, thus this is true.
Now for step 2. Do I continue with "Suppose $n\geqslant k>2$" ? I don't see where to go next. Can anyone help?