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

06 Dec, 2021
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++
Java
Python
C#
PHP
Javascript
// 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 Codeint main(){    int n = 10;    cout << Square_Diff(n);    return 0;     } // This code is contributed by 'Gitanjali' .
// 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 Codepublic static void main(String s[]){    int n = 10;    System.out.println(Square_Diff(n));         }}// This code is contributed by 'Gitanjali'.
# 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 codeprint(Square_Diff(10))
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 program to find the difference// between sum of the squares of the// first n natural numbers and square// of sum of first n natural numberfunction 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 .?>
<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 numberfunction 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 Codevar n = 10;document.write(Square_Diff(n)); // This code is contributed by Princi Singh</script>
Output : 
2640
 
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