The sum of the first n odd numbers is always a square

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 n odd numbers is always a square. For example,
11 + 3 = 41 + 3 + 5 = 91 + 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) = 1sum_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 :-):

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?

Unknown said…
Just noticing that they agree on the first three values

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.
James said…
What do you do on long flights?
Watch movies...
turd said…
The sum of the first n odd numbers is = to n^2. Im in 7th grade and I know that
turd said…
the sum of the first n odd numbers is n^2. U ok sir im in 7th grade and i know that
turd said…
The sum of the first n odd numbers is = to n^2. Im in 7th grade and I know that
Unknown said…
Find the sum of first n odd numbers............1,3,5,7,.........2n-1........will you please give me the answer????? I am 10 std student
Unknown said…
Write a C program to take the lower limit and upper limit of the range as inputs and calculate!

i. Square sum of those interval numbers
ii. Show the odd numbers only within the interval
iii. Determine the average of those numbers.
Unknown said…
This comment has been removed by the author.
Unknown said…
Just to make it more fun
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
Unknown said…
Please answer: the sum of odd numbers from 10 to 60 is??
Unknown said…
25/2*(11+59)
Since 25 odds from 10 to 60
=875

Your last name contains invalid characters

My last name is "Graham-Cumming". But here's a typical form response when I enter it:

Does the web site have any idea how rude it is to claim that my last name contains invalid characters? Clearly not. What they actually meant is: our web site will not accept that hyphen in your last name. But do they say that? No, of course not. They decide to shove in my face the claim that there's something wrong with my name.

There's nothing wrong with my name, just as there's nothing wrong with someone whose first name is Jean-Marie, or someone whose last name is O'Reilly.

What is wrong is that way this is being handled. If the system can't cope with non-letters and spaces it needs to say that. How about the following error message:

Our system is unable to process last names that contain non-letters, please replace them with spaces.

Don't blame me for having a last name that your system doesn't like, whose fault is that? Saying "Your last name …

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You need: Java 6 and openssl.

1. Suppose you have a certificate and key in PEM format. The key is named host.key and the certificate host.crt.

2. The first step is to convert them into a single PKCS12 file using the command: openssl pkcs12 -export -in host.crt -inkey host.key > host.p12. You will be asked for various passwords (the password to access the key (if set) and then the password for the PKCS12 file being created).

3. Then import the PKCS12 file into a keystore using the command: keytool -importkeystore -srckeystore host.p12 -destkeystore host.jks -srcstoretype pkcs12. You now have a keystore named host.jks containing the certificate/key you need.

For the sake of completeness here's the output of a full session I performe…

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As further proof of my unsuitability to be a child minder (see previous post) I found myself playing with an Ikea LILLABO 20-piece basic set train.

The train set has 16 pieces of track (12 curves, two straight pieces and a two part bridge) and 4 pieces of train. What I wondered was... how many possible looping train tracks can be made using all 16 pieces?

The answer is... 9. Here's a picture of the 9 different layouts.

The picture was generated using a little program written in Processing. The bridge is red, the straight pieces are green and the curves are blue or magenta depending on whether they are oriented clockwise or anticlockwise. The curved pieces can be oriented in either way.

To generate those layouts I wrote a small program which runs through all the possible layouts and determines which form a loop. The program eliminates duplicate layouts (such as those that are mirror images of each other).

It outputs a list of instructions for building loops. These instructions con…