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# 1. Prove the binomial theorem

Recall that if F(z) = ∑ akzk, then we can extract ak by differentiating F k times, dividing by k!, and setting the result to 0, i.e. ak = (1/k!) dk/dzk F(z) |z=0.

1. Prove by induction on k that dk/dzk (1+z)n = n(k) (1+z)n-k for all k in ℕ.

2. Use this fact to show that if (1+z)n = ∑ akzk, then ak = n(k)/k!.

3. Use the preceding to prove that (x+y)n = ∑ (n(k)/k!) xkyn-k.

# 2. A triple identity

Recall Vandermonde's identity (see BinomialCoefficients):

A combinatorial interpretation of this identity is that if you want to pick k things out of the union of an n-element set and an m-element set, you can do so by first picking a value r, and then choosing r things from the first set and k-r from the second.

Consider the following alleged identity, which we will call the triple Vandermonde identity or TVI for short:

1. Give a combinatorial interpretation of TVI.
2. Prove TVI is true.

# 3. Sequence differences

1. Suppose you have a generating function F(z) = ∑ anzn. Write an expression in terms of F for the generating function G of the sequence given by the rule bn = an-an-1.

2. Prove the identity

Note: Typo in the identity corrected 2007-10-08; previous version had the wrong operation on the right-hand side.

# 4. Rabbits

Let's imagine that Fibonacci's famous rabbits are so fecund that they produce two daughters in their second year instead of just one.

Specifically, let T(n) satisfy the recurrence T(n) = T(n-1) + 2T(n-2), with T(1) = T(0) = 1. Find a closed-form expression for T(n).

2014-06-17 11:57