Why is the sum of the first n odd numbers the square of n?

Answer 1

1+3+5+7+9+...........+(2n-1) where n=1,2,3,4,....

In AP first term a=1 and diff d=2

As we know in AP (d-a)+(2d-a)+.........+(nd-a) if d>a and n=1,2,3,..
sum= n(n+1)d/2 -na where n=1,2,3..
substituting a=1 and d=2 ( for odd numbers)
sum=n^2+n-n
sum=n^2.

Answer 2

n^2 = n + n + n + ... +n (n times)

= (1 + n-1) + (2 + n-2) + ...+ (n-1 + (n-(n-1))) + (n + (n-n))

Rearrange

= 1 +(n-n) + 2 + (n-(n-1)) + 3 + (n - (n-2) + ... n +( n -1)

= 1 + (2 + 1) + (3+2) + (4+3) + ... + 2n-1

= 1 + 3 + 5 + ... + 2n-1 = sum of first n odd numbers

You can see this graphically if you take a square checkerboard, cut it the edges of the squares that are on the diagonal.
You get two pieces , one that has rows of 1,2,3,4,...n squares,
another with rows with 1,2,3,...(n-1) squares. Drop the small piece on top of the large piece so the corners line up and count the combined squares in each row. The will be 1,3,5, 7...

Answer 3

Because (n +1)**2 = n**2 + 2n +1.

Note, 2n + 1 is always odd, because 2n is always even.

So to get to the next square, you add the next odd number to it.

as in (3 + 1)**2 = 3**2 + 3*2 + 1

Adding up the odd numbers has to be odd squares because of that.

It is possible to prove that with a series sumnation, but there are too many symbols involved to post it here. Yahoo does not do sigma signs well.

Now, a more interesting question.

The equation a**2 + b**2 = c**2 is solvable in integers (that would be the sides of a right triangle)

The first one is

3 4 5 (9 + 16 = 25)

Using what you just learned about the sum of odd numbers, can you come up with an infinite number of linearly independent solutions to that problem?

Hint:

The next two are

5 12 13
7 24 25

Drop me a line at [e-mail] if you want the full answer.

-Dio

Answer 4

Draw an nxn square:

****
****
****
****

Now, starting from the lower left corner, mark off successively larger squares. This is difficult to do in ascii, but you should have n squares, starting from a 1x1 and increasing up to an nxn. Now look at the L-shaped area between the squares. The first L has 3 stars, the second has 5, the third has 7, etc. Since th sum of 1 (the first square) and all the L-shapes (3+5+7+..) gives the large square n^2, you get your result.

Answer 5

Oooo! An inductive proof! Cool!

I don't suppose I have to prove that the nth odd number is 2n - 1, do I?

Base Case: Let n = 1. The sum of the first 1 odd numbers is clearly 1. 1 squared is clearly 1. The base case is true.

Induction Step: Suppose the statement is true for some n ≥ 1.
n
Σ (2i-1) = 1 + 3 + ... + (2n - 1) = n^2.
i=1
Thus (n+1)^2 = n^2 + 2n + 1
. . . . . . . . . . . . . = (1 + 3 + ... + (2n - 1)) + (2n + 1)
. . . . . . . . . . . . . = (1 + 3 + ... + (2n - 1)) + ((2(n+1) - 2) + 1)
. . . . . . . . . . . . . = (1 + 3 + ... + (2n - 1)) + (2(n+1) - 1)
. . . . . . . . . . . . . . n+1
. . . . . . . . . . . . . = Σ (2i-1)
. . . . . . . . . . . . . . i=1
which is the sum of the first n+1 odd numbers.

QED

Answer 6

True enough. Oddity of math is one way of putting it. Perhaps a more informative way would be to say that it is mathematically provable. It is easy to show by induction.

n=1, n^2=1. So it is true for n=1.

Assume for some n, n^2=1+3+...+(2n-1).
We must show that (n+1)^2=1+3+...+(2n-1)+(2(n+1)-1).

Add the (n+1)st odd number to the first expression:
n^2 + (2(n+1)-1) = 1+3+...+(2n-1)+(2(n+1)-1)

Simplify:
n^2+2n+1 = 1+3+...+(2n-1)+(2n+1)

Factor the left hand side:
(n+1)^2 = 1+3+...+(2n-1)+(2n+1), proving the equality needed.

Answer 7

just an oddity of math