2. Consider a complete d-ary tree with height h, that is, a tree where
the root is at level 0, leaves are at height h, and all internal nodes
have d children (i.e, every level in the tree is full except the last
level).

a. What is the upper bound on the numbers of leaves in such a tree? 

b. Find lower and upper bounds for the total number of nodes

Consider as a special case a 3-ary tree.  The first few levels look
like this:
                       o          Level 0
                      /|\
                     / | \
                    /  |  \
                   o   o   o      Level 1
                  /|\ /|\ /|\
                  ooo ooo ooo     Level 2

As you can see, the first level contains 1 node.  The second level
contains 3 nodes.  The third level contains 9 nodes.  The pattern
continues.  For a d-ary tree, the nth level contains d^n nodes. 
Therefore, the first through nth level contain d^0 + d^1 + d^2 + . . .
+ d^n nodes.

With that background, on to your questions:

a)   Level h contains at most d^h nodes.  If h is the height, these
nodes are leaves.

b)   With height h, we know there are h-1 fully-populated levels. 
Level h must have at least one node and may have at most d^h nodes. 
So, the lower bound on the total number of nodes is d^0 + d^1 + d^2 +
. . . + d^(h-1) + 1; the upper bound is d^0 + d^1 + d^2 + . . . +
d^(h-1) + d^h.

Does that make sense with the picture drawn out?

-Haversian