I was staring at the checked pattern on the back of an airline seat the other day when I suddenly saw that the sum of the first

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

So, the sum of the first

But using standard formulae is annoying, so how about trying a little induction.

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

And given that we know from Gauss the sum of the first

What do you do on long flights?

`n`odd numbers is always a square. For example,

1

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)

= n^2

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

= (n+1)^2

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

= n^2

What do you do on long flights?

## Comments

1 = 1

1 + 3 = 4

1 + 3 + 5 = 9

is actually a valid proof. $\sum_{i=1}^n p(i)$, where $p(i)$ is a polynomial of degree $d$, is a polynomial of degree at most $d + 1$ (you can prove this by induction). Now $n^2$ and $\sum_{i=1}^n (2n -1)$ are both polynomials of degree at most 2, and they agree on three values, hence must be identical.

Watch movies...

i. Square sum of those interval numbers

ii. Show the odd numbers only within the interval

iii. Determine the average of those numbers.

1=1^3

3+5=2^3

7+9+11=3^3

Proof isn't quite so pretty, but you can notice their average is always n^2

Since 25 odds from 10 to 60

=875