Strange pattern

Look at the sums of consecutive odd numbers, starting from 1:
1 = 1
1+3 = 4
1+3+5 = 9
1+3+5+7 = 16
1+3+5+7+9 = 25
1+3+5+7+9+11 = 36
...and so on.

Now look at consecutive perfect squares:
1x1 = 1
2x2 = 4
3x3 = 9
4x4 = 16
5x5 = 25
6x6 = 36

Is there any good mathmatical reason for this?  Is there any way to prove it?  My father claims to have discovered this on his own, and I believe him, but there's no way he could have been the first.  I've looked at various websites but I've never seen anyone else talk about this pattern.  Is there even a name for it?

2get

this is the easiest induction proof you will ever do.  observe what you're claiming: the sum of the first n positive odd numbers is equal to n².  In order to prove that this is true for all natural numbers ('ordinary' numbers from 1,2,3.. on up), remarkably, all you have to do is prove two things: the assertion is true for the base case (1), and if the statement being true for an arbitrary larger number n implies that the statement is true for the next case, n+1, the statement is true for all natural n.  This is called the Principal of Mathematical Induction, an extremely useful proof technique, and you've happened upon basically the single easiest example of it.  First of all, clearly, the first one positive integer, 1, is equal to 1², so our base case is established.  now to do the inductive step, we need a general form for odd numbers.  1=(2(1)-1), 3=(2(2)-1), 5=(2(3)-1) and so forth.  I will leave it as an exercise, as the math books say, to show that the nth odd positive integer is given by 2n-1.  I will assume it for our formula.  Now what is the statement about n for arbitrary n?  It is this equality:  1+3+5+7+...+(2n-1)=n²  Again, the first n odd positives sum to n².  We are allowed to ASSUME that this equation is true (this is called the inductive hypothesis); what we have to do now, is, given that assumption, show that the truth of the equation holds for the next case, n+1.  So, adding the next odd positive number to the left hand side of the equation should make that equal to (n+1)².  The next odd positive number is 2n+1 (add two).

So,  1+3+5+7+...+(2n-1)+(2n+1) = (1+3+5+7+...+(2n-1)) + (2n+1)
     = n² + (2n+1).  We are allowed to substitute n² for most of the sum, again, by the inductive hypothesis: we can assume that they are equal.

     Now, n² + 2n + 1 = (n+1)² (just factor).  so, the long sum from above is equal to the square of the next natural number, exactly what we set out to prove.  this establishes the observation for all natural n.  Nor is this somehow 'circular' reasoning, as some students might worry: all we're saying is that a certain condition holds on certain restraints (formulae) for larger and larger n.

THE JEWS CREATED IT

Awesome, thanks for the explanation.  To be honest, I never even bothered taking a crack at it myself (I'm more of a language guy, not a math guy).

By the way, I see you use the name "York" -- I had a math teacher back in highschool with that last name.  Any chance you're middle-aged bald guy who lives in southern California, or are all Yorks just coincidentally good with math?

Wouldn't it be easier to use the arithmetic sum formula:

SUM 1-->n (a) = n * (a(1) + a(n)) / 2

therefore...

SUM 1-->n (2i - 1)
= n * (2*1-1 + 2n-1) / 2
= n * 2n / 2
= n^2

Even easier

S= 1 + 3 + 5 + 7 + ... + (2n-1)

So 2S = (1 + 2n-1) + (3 + 2n-3) + (5 + 2n-5) + ...
      = 2n + 2n + 2n + ... (n times)
      = 2 n^2

So S = n^2

This "reverse the sequence and add" thing is how you derive the arithmetic sum formula, I think.

so if i know the square of say 13, i can find the square of 14 without multiplying!

13² = 169
(n + 1)²  = n² + 2n + 1
14² = 169 + 2(13) + 1 = 196

sweeet! i never thought of it like that before!

>Is there even a name for it?

Pascal's Triangle?

The most simple anser is to look at the average number, which is about half of the highest number.  So you have your number of numbers in there as part of the average, but you have to multiply that my the total number of numbers again to get a real sum.  So you're multiplying something by itself.  The details are a little more complicated, but this is the basic reason you end up seeing a square series.

<Inst>
um, because distance between integer squares is

(X+1)^2 - X^2
X^2+2X+1-X^2
contract...
2X+1.

<Inst>
Okay, wording is completely wrong, but it works am i rite?

X= square root of perfect square.

ok how about this

1^3 = 1
1^3+3^3 = 1+27=28
1^3+3^3+5^3+7^3= 1+27+125+343 = 496
1^3+3^3+5^3+7^3+9^3+11^3+13^3+15^3 = 1+27+125+343+729+1331+2197+3375 = 8128

also
divisors of 1 are 1 and 1=1
divisors of 28 are 1,2,4,7,14 and 1+2+4+7+14=28
divisors of 496 are 1,2,4,8,16,31,62,124,248 and 1+2+4+8+16+31+62+124+248=496
divisors of 8128 are 1,2,4,8,16,32,64,127,254,508,1016,2032,4064 and 1+2+4+8+16+32+64+127+254+508+1016+2032+4064=8128

...

I never knew this construction for perfect numbers.  It might be very useful in settling whether or not there are odd perfect numbers.

meh, i doubt it.

bump