[Remember Your Squares] Something I Found

[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");
}

 

[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

 

[SomeGuy121]

Are you being sarcastic with "congratulations"?