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)
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