T minus 100

It’s 100 days until my birthday! (And how cool that the hundredth day before my birthday is Bastille Day! Well, it’s a little cool. Oh, OK, it’s a meaningless coincidence.)

To goose my weight-loss regime, which appears to have stalled once again — though I am holding my own against the “Google 15” — I am adding a daily exercise regimen for the first time, inspired by hundredpushups.com: each day I will do one more push-up than the day before, starting with one today and culminating with a hundred push-ups by my birthday.

If I keep exactly to that plan, the total number of push-ups I’ll do is 5,050 — one today, two tomorrow, three on Wednesday, four on Thursday, and so on. The sum of the first N numbers from 1 through N is, in general,

(N+1) × N/2

an elegant intuitive proof of which is as follows. List the first N numbers, let’s say 6 for this example:

1 2 3 4 5 6

The sum of the two “outer” numbers is 7:

1 2 3 4 5 6

Removing those, the sum of the next two “outer” numbers is also 7:

2 3 4 5

Removing those, the sum of the final pair is also 7:

3 4

That sum — N+1 — is repeated N/2 times, giving rise to the formula

(N+1) × N/2

“Wait a minute,” I hear you say. “What about when N is odd? Then there’s one extra innermost number with no partner.” That’s true. In that case, the number of pairs that add up to N+1 isn’t N/2, it’s only (N-1)/2:

1 2 3 4 5 6 7

Here N is 7, and there are 3 pairs that add up to 8 — 1 and 7, 2 and 6, 3 and 5 — and 4 is all alone in middle. So the sum is:

(N+1) × (N-1)/2 + the middle number

But the middle number is always (N+1)/2, so this becomes:

(N+1) × (N-1)/2 + (N+1)/2

which is the same as

(N+1)/2 × (N-1) + (N+1)/2

which can be read as adding one more (N+1)/2 to a collection of N-1 of them, for a total of N (N+1)/2’s:

(N+1)/2 × N

which is the same as

(N+1) × N/2

which is the same as the original formula above whether N is odd or even. QED.

OK, let’s get this regimen started. Rrrrrnnnnngghh — one. Whew.

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