# [Remember Your Squares] Something I Found

1. Feb 17, 2010

### SomeGuy121

Let x = 1.
Let n = Next Odd Number
Let y = Previous Sum

x2 = x
+3 = 4 = (1+x)2
+5 = 9 = (3+x)2
+n = n+y = (n-2 + x)2

You could make a program to list all the squares without invoking the multiplication function or squaring function using a simple loop.

C++ Example:

#include "stdafx.h"
#include <iostream>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
cout<<"Squares: \n\n";
int sum=1,nextOdd=1; // Sum is Starting Integer Squared, Declared X in the For Loop below

for(int x=1;x<100;x++)
{
cout<<x<<" Squared is "<<sum<<"\n";
nextOdd+=2;
sum+=nextOdd;
}
system("pause");
}

2. Feb 17, 2010

### CRGreathouse

You've discovered that
$$\sum_{k=1}^n2k-1=n^2$$.
Congratulations.

Can you transform that into a formula for $$\sum_{k=1}^nk$$? Can you find one for $$\sum_{k=1}^nk^2$$?

You can check your work afterward, and even glimpse what's beyond:
http://mathworld.wolfram.com/FaulhabersFormula.html

3. Feb 17, 2010

### SomeGuy121

Are you being sarcastic with "congratulations"?