前 n 个自然数的平方和与和的平方之差

给定一个整数 n，求前 n 个自然数的平方和与前 n 个自然数的平方和的绝对差。
例子：

Input : n = 3
Output : 22.0
Sum of first three numbers is 3 + 2 + 1 = 6
Square of the sum =  36
Sum of squares of first three is 9 + 4 + 1 = 14
Absolute difference = 36 - 14 = 22

Input : n = 10
Output : 2640.0

询问于：biwhiz 公司

方法：
1. 求前 n 个自然数的平方和。
2. 求前 n 个数之和，然后平方。
3. 找出两者总和之间的绝对差值并打印出来。
以下是上述方法的实现：

C++

// C++ program to find the difference
// between sum of the squares of the
// first n natural numbers and square
// of sum of first n natural number
#include 
using namespace std;
 
int Square_Diff(int n){
 
int l, k, m;
 
    // Sum of the squares of the
    // first n natural numbers is
    l = (n * (n + 1) * (2 * n + 1)) / 6;
     
    // Sum of n naturals numbers
    k = (n * (n + 1)) / 2;
 
    // Square of k
    k = k * k;
     
    // Differences between l and k
    m = abs(l - k);
     
    return m;
 
}
 
// Driver Code
int main()
{
    int n = 10;
    cout << Square_Diff(n);
    return 0;
     
}
 
// This code is contributed by 'Gitanjali' .

Java

// Java program to find the difference
// between sum of the squares of the
// first n natural numbers and square
// of sum of first n natural number
 
public class GfG{
 
static int Square_Diff(int n){
 
int l, k, m;
    // Sum of the squares of the
    // first n natural numbers is
    l = (n * (n + 1) * (2 * n + 1)) / 6;
     
    // Sum of n naturals numbers
    k = (n * (n + 1)) / 2;
 
    // Square of k
    k = k * k;
     
    // Differences between l and k
    m = Math.abs(l - k);
     
    return m;
 
}
 
// Driver Code
public static void main(String s[])
{
    int n = 10;
    System.out.println(Square_Diff(n));    
     
}
}
// This code is contributed by 'Gitanjali'.

Python

# Python3 program to find the difference
# between sum of the squares of the
# first n natural numbers and square
# of sum of first n natural number
 
def Square_Diff(n):
 
    # sum of the squares of the
    # first n natural numbers is
    l = (n * (n + 1) * (2 * n + 1)) / 6
     
    # sum of n naturals numbers
    k = (n * (n + 1)) / 2
 
    # square of k
    k = k ** 2
     
    # Differences between l and k
    m = abs(l - k)
     
    return m
 
# Driver code
print(Square_Diff(10))

C#

using System;
 
public class GFG
{
 
  static int Square_Diff(int n)
  {
 
    int l, k, m;
 
    // Sum of the squares of the
    // first n natural numbers is
    l = (n * (n + 1) * (2 * n + 1)) / 6;
 
    // Sum of n naturals numbers
    k = (n * (n + 1)) / 2;
 
    // Square of k
    k = k * k;
 
    // Differences between l and k
    m = Math.Abs(l - k);
 
    return m;
 
  }
 
  // Driver Code
  public static void Main()
  {
    int n = 10;
    Console.WriteLine(Square_Diff(n));   
  }
}
 
// This code is contributed by akshitsaxena09.

PHP

Javascript

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