A reader wrote in with a great problem. This was months ago.

Wargle is an Irish blogger. Math and politics. Not the same as jd2718, but something like an Irish cousin… definitely worth a look.

The math tends to be more advanced than mine. Much of the politics is EU. Plus some random stuff. And language. And kids.

Anyhow, he wrote me privately, said he had this great problem, and I agreed. And I was going to use it to teach geometric sequences. Use it as my intro… But I forgot. Here’s how I introduced series, instead. That evening, I remembered Wargle/tax and kicked myself.

So the next day, now we’ve covered all of our 5 series and sequence topics, they kidlets have their desks lined up, ready to quiz, and I say to keep their notebooks out, and I dictate a(n Americanized version of) Wargle’s problem.

Man, they worked hard!  And in both classes, the same result:  some kids completed the list of numbers and summed them. Some kids guessed towards the sum, and weren’t sure why, some kids recognized the sequence, and used our formula, and one kid in each class suggested that the Total minus 40% of the Total should be \$1000 (no euros, sorry).

What a great coming together, satisfying for all. I am glad I used the question on the second day rather than the first.

Anyhow, here’s Wargle’s version:

### Proof by taxation

A little story

One day your boss says, “Great job, we want to give you a bonus, were you planning a holiday this year?”. You say “well actually, I wanted to go here”, you show him a website, “but I can’t afford it”. “How much is it?”. “1000 euro”, you say. “Great, we’ll pay for that”. “Thanks boss!”. Happy days.

Later you realise that if they just give you 1000e you’ll have to pay tax on that. The tax rate is 40% so that’s a 400e tax bill, you still can’t afford the holiday.

You head up to your boss’s office and explain the problem. He’s still in a great mood and says, “Don’t worry, we’ll pay your tax bill too”. “Wow but there’s another bit of a problem, I’ll have to pay tax on that extra 400 too. 40% of 400e is 160e, I still can’t afford it”. You boss is looking less happy now and gets out a piece of paper. “Right, we’ll pay all your tax, no matter what” and starts writing down some figures

1000
400
160
64
25.60
10.24
4.096
1.6384
0.65536

you hear him mumbling as he starts adding them all up. Suddenly Joan, his secretary, who’s been quiet all this time blurts out “1666e and 66c”. You both stare at her in amazement then after a lot more mumbling the boss says “my total is 1666e and .23c but I suppose if I added a few more lines to the sum it’d probably be 1666.66. How did you get it Joan?”

“Well, tax is 40% so he gets to keep 60% of anything you pay him. So the final amount in his pocket is the what you give him times .6 . So you’re looking for a number that when you multiply it by .6 gives you 1000. So 1000 ÷ .6 is the number you’re looking for because when you multiply that by .6 the two .6s cancel out and only the 1000 is left and 1000 ÷ .6 is 1666.666666666…”, says Joan.

So what we have proved is that 1000 + 1000 x .4 + 1000 x .4^2 + 1000 x .4^3 + … = 1000 ÷ .6 . There was nothing special about our choice of 40%, so replacing .4 with r (and .6 by 1-r) and dividing out the 1000 on both sides gives 1 + r + r^2 + r^3 + … = 1 ÷ (1 – r) Of course there are other ways to prove this but, I like my proof by taxation because it feels like it explains why they are equal.

1. March 19, 2010 am31 10:06 am 10:06 am

I love it! So simple, yet it makes so much sense.

2. March 20, 2010 am31 11:18 am 11:18 am

Tasty and delicious.

3. March 22, 2010 pm31 1:17 pm 1:17 pm

What a great problem! I’ve noticed as I’ve taught about geometric series that most (perhaps all?) questions that have a natural context can be solved by a nice observation (+ basic algebra) instead of mindlessly computing $\frac{a}{1-r}$. I’ll give a couple of examples in a second, but what I love about this problem is how general it is. The tax could be any percent greater than 0 and less than 100 (and it is clear that if tax were at least 100%, then no bonus could possibly be enough), and the cost of vacation could be any positive amount, and similar reason would show that the formula we teach is absolutely correct (and MAKES SENSE!).

Here are a couple of geometric series questions I’ve seen that have such slick solutions:

Abby and Bob decide to play a game. They will take turns throwing a fair coin until one of them gets a heads. That person is declared the winner. Bob decides to let Abby go first (probably because her name starts with the first letter of the alphabet). What is the probability that Abby wins the game?

The long solution involves finding the probability that Abby wins in round n and then summing this (geometric) series. The short way is to recognize that the odds that Abby wins in the first round are twice the odds that Bob wins in the first round (1/2 to 1/4). There is nothing special about the first round, however: If the game makes it to round n, Abby is twice as likely to win in that round as Bob. So the total probability that Abby wins must be twice the total probability that Bob wins. Since the probability that the game goes on forever is 0, we can conclude that the total probability that Abby wins the game is 2/3 (which is in 2:1 ratio to the 1/3 probability that Bob wins the game).

A similar trick works for the type of problems that asks to find a certain area in a drawing with nested figures. It is enough to look at what happens at a single “step” in the nest and compute the ratio of shaded to unshaded in this “step.” Because of the way the nest is set up, each step will give this same ratio. This then allows you to compute the total area of the shaded region.

Probably the most famous “geometric series” question involves two trains on a collision course. The initial distance between the trains is known, as are their respective velocities. A bird (or fly or bee) flies back and forth between the trains until it is eventually squashed by the collision. The birds velocity is also known and we assume it reverses direction instantaneously. The question is how far the bird flies before its untimely (or timely) death. The long solution involves finding the length of each leg of the birds trip and then summing this (geometric) series. The short way is to just find out how long it will take for the trains to collide and multiply that by the speed of the bird.

Thanks for the great problem, and sorry for such a long comment.