Combinatorics Through Guided Discovery, 2004a

190 D. Hints to Selected Problems
349. To show a relation is an equivalence relation, you need to show it satisfies the
definition of an equivalence relation.
353. To get you started, in Problem 38 the equivalence classes correspond to seating
arrangements.
361. You’ve probably guessed that the sum is n 2 . To prove this by contradiction,
you have to assume it is false, that is, that there is an n such that 1+3+
5+···+2n − 1 n 2 . Then the method of Problem 360 says there must
be a smallest such n and suggests we call it k. Why do you know that
1+3+5+···+2k − 3=(k − 1)2? What happens if you add 2n − 1 to both
sides of the equation?
365. You’ve probably already seen that, with small values of n, sometimes n 2 and
sometimes 2 n is bigger. But if you keep experimenting one of the functions
seems to get bigger and stay bigger than the other. The number n = b where
this change occurs is a good choice for a base case. So as not to spoil the
problem for you, we won’t say here what this value of b is. However you
shouldn’t be surprised later in the proof if you need to use the assumption
that n/gtb.
Additional Hint: You may have reached the point of assuming that 2 k−1 >
(k − 1) 2 and found yourself wondering how to prove that 2 k > k 2 . A natural
thing to try is multiplying both sides of 2 k−1 > (k − 1) 2 by 2. This ends up
giving you 2 k > 2k 2 − 4k +2. Based on previous experience it is natural for
you to expect to see how to turn this new right hand side into k 2 but not see
how to do it. Here is the hint. You only need to show that the right hand side
is greater than or equal to k 2 . For this purpose you need to show that one of
the two k 2 sin2k 2 somehow balances out the −4k. See if you can figure out
how the fact that you are only considering ks with k > b can help you out.
366. When you suspect an argument is not valid, it may be helpful to explicitly try
several values of n to see if it makes sense for them. Often small values of n
are adequate to find the flaw. If you find one flaw, it invalidates everything
that comes afterwards (unless, of course, you can fix the flaw).
370. You might start out by ignoring the word unique and give a proof of the
simpler theorem that results. Then look at your proof to see how you can
include uniqueness in it.
377. An earlier problem may help you put your answer into a simpler form.
378. What is the power series representation of e x2 ?
381. There is only one element that you may choose. In the first case you either
choose it or you don’t.
387.b. At some point, you may find the binomial theorem to be useful.