Difference between sum of the squares of first n natural numbers and square of sum

Given an integer n, find the absolute difference between sum of the squares of first n natural numbers and square of sum of first n natural numbers.
Examples : 

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

Asked in : biwhiz Company 

Approach :
1. Find the sum of square of first n natural numbers. 
2. Find the sum of first n numbers and square it. 
3. Find the absolute difference between both the sums and print it.
Below is the implementation of above approach : 

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 <bits/stdc++.h>
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

<?php

// PHP 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
function Square_Diff($n)
{

    $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
    $n = 10;
    echo Square_Diff($n);

// This code is contributed by anuj_67 .
?>

Javascript

<script>
// javascript 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
function Square_Diff(n){

var 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
var n = 10;
document.write(Square_Diff(n));

// This code is contributed by Princi Singh
</script>

Output : 

2640

Time Complexity: O(1)
Auxiliary Space: O(1)