Sum of squares and peace parties

sum of squares and peace parties Problem Description:

The sum of squares of the first 10 natural numbers is 12+22+32+...+102=385, the first 10 natural numbers and the square is (1+2+3+...+10) 2=552=3025;

The difference between the sum of squares and the sum of squares is 3025-385=2640. Now given an n, ask you to find the sum of the squares of the first n natural numbers and the difference between the peace and the square.

Input:

An integer n

Output:

An integer (the sum of the squares of the first n natural numbers and the difference between the peace Parties)

For example

Input:

10

Output:

2640 Solution One:

The first and the most stupid solution, that is, directly to the problem of the calculation process simulation to obtain,,, the final direct output result

The C language code is as follows:

#include <stdio.h>
#include <math.h>
void Main ()
{
    __int64 result,sum1=0,sum2=0;	
    int n,i;	
    scanf ("%d", &n);	
    for (i=1;i<=n;i++)		
    {		
        sum1+=pow (i,2);		
        sum2+=i;		
    }	
    Sum2=pow (sum2,2);	
    RESULT=SUM2-SUM1;
    printf ("%i64d\n", result);
}

Solution Two:

First of all, let's write down the two formulas in the question first.

The Final Solution is:

The C language code is as follows:

#include <stdio.h>
void Main ()
{
	__int64 result=0;
	int n,i,j;
	scanf ("%d", &n);
	for (i=1;i<n;i++)
	{
		result+=2*i* (((n-i) * (i+1+n))/2);
	}
	printf ("%i64d\n", result);
}

Solution Three:

This formula is proved by mathematical inductive method

The C language code is as follows:

#include <stdio.h>
#include <math.h>
void Main ()
{
	__int64 result,n,sum1,sum2;
	scanf ("%i64d", &n);
	Sum1= (n (n+1) * (2*n+1))/6;
	Sum2=pow (n (n+1))/2,2);
	RESULT=SUM2-SUM1;
	printf ("%i64d\n", result);
}