1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
And, of course, it occurred to me that it would be nice to be able to prove it. There are lots of ways to do that. Firstly, this is just the sum of an arithmetic progression starting at a = 1 with a difference of d = 2. So the standard formula gives us:
sum_odd(n) = n(2a + (n-1)d)/2
= n(2 + (n-1)2)/2
= n(1 + n - 1)
So, the sum of the first n odd numbers is n^2.
But using standard formulae is annoying, so how about trying a little induction.
sum_odd(1) = 1
sum_odd(n+1) = sum_odd(n) + (2n + 1)
= n^2 + 2n + 1
But back to the airline seat. Here's what I saw (I added the numbering, Lufthansa isn't kind enough to do that for you :-):
The other thing I noticed was this:
You can view the square as the sum of two simpler progressions (the sum of the first n numbers and the sum of the first n-1 numbers):
1 + 3 + 5 + 7 =
1 + 2 + 3 + 4 +
1 + 2 + 3
And given that we know from Gauss the sum of the first n numbers if n(n+1)/2 we can easily calculate:
sum_odd(n) = sum(n) + sum(n-1)
= n(n+1)/2 + (n-1)n/2
= (n^2 + n + n^2 - n)/2
What do you do on long flights?