Given a positive integer n. The task is to find the sum of the sum of square of first n natural number.
Examples :
Input : n = 3 Output : 20 Sum of square of first natural number = 1 Sum of square of first two natural number = 1^2 + 2^2 = 5 Sum of square of first three natural number = 1^2 + 2^2 + 3^2 = 14 Sum of sum of square of first three natural number = 1 + 5 + 14 = 20 Input : n = 2 Output : 6
Method 1: O(n) The idea is to find sum of square of first i natural number, where 1 <= i <= n. And then add them all.
We can find sum of squares of first n natural number by formula: n * (n + 1)* (2*n + 1)/6
Below is the implementation of this approach:
C++
// CPP Program to find the sum of sum of // squares of first n natural number#include <bits/stdc++.h>using namespace std;// Function to find sum of sum of square of // first n natural numberint findSum(int n){ int sum = 0; for (int i = 1; i <= n; i++) sum += ((i * (i + 1) * (2 * i + 1)) / 6); return sum;}// Driven Programint main(){ int n = 3; cout << findSum(n) << endl; return 0;} |
Java
// Java Program to find the sum of // sum of squares of first n natural// numberclass GFG { // Function to find sum of sum of // square of first n natural number static int findSum(int n) { int sum = 0; for (int i = 1; i <= n; i++) sum += ((i * (i + 1) * (2 * i + 1)) / 6); return sum; } // Driver Program public static void main(String[] args) { int n = 3; System.out.println( findSum(n)); }}// This code is contributed by // Arnab Kundu |
Python3
# Python3 Program to find the sum# of sum of squares of first n # natural number# Function to find sum of sum of# square of first n natural numberdef findSum(n): summ = 0 for i in range(1, n+1): summ = (summ + ((i * (i + 1) * (2 * i + 1)) / 6)) return summ # Driven Programn = 3print(int(findSum(n)))# This code is contributed by # Prasad Kshirsagar |
C#
// C# Program to find the sum of sum of // squares of first n natural numberusing System;public class GFG { // Function to find sum of sum of // square of first n natural number static int findSum(int n) { int sum = 0; for (int i = 1; i <= n; i++) sum += ((i * (i + 1) * (2 * i + 1)) / 6); return sum; } // Driver Program static public void Main() { int n = 3; Console.WriteLine(findSum(n)); }}// This code is contributed by// Arnab Kundu. |
PHP
<?php// PHP Program to find the sum of // squares of first n natural number// Function to find sum of sum of // square of first n natural numberfunction findSum( $n){ $sum = 0; for ($i = 1; $i <= $n; $i++) $sum += (($i * ($i + 1) * (2 * $i + 1)) / 6); return $sum;}// Driver Code$n = 3;echo findSum($n) ;// This code is contributed by anuj_67.?> |
Javascript
<script>// Javascript Program to find the sum of sum of // squares of first n natural number // Function to find sum of sum of square // of first n natural number function findSum(n) { return (n * (n + 1) * (n + 1) * (n + 2)) / 12; } // Driven Program let n = 3; document.write(findSum(n) + "<br>"); // This code is contributed by Mayank Tyagi</script> |
Output :
20
Time Complexity : O(n)
Auxiliary Space : O(1)
Method 2: O(1)
Mathematically, we need to find, ? ((i * (i + 1) * (2*i + 1)/6), where 1 <= i <= n
So, lets solve this summation,
Sum = ? ((i * (i + 1) * (2*i + 1)/6), where 1 <= i <= n
= (1/6)*(? ((i * (i + 1) * (2*i + 1)))
= (1/6)*(? ((i2 + i) * (2*i + 1))
= (1/6)*(? ((2*i3 + 3*i2 + i))
= (1/6)*(? 2*i3 + ? 3*i2 + ? i)
= (1/6)*(2*? i3 + 3*? i2 + ? i)
= (1/6)*(2*(i*(i + 1)/2)2 + 3*(i * (i + 1) * (2*i + 1))/6 + i * (i + 1)/2)
= (1/6)*(i * (i + 1))((i*(i + 1)/2) + ((2*i + 1)/2) + 1/2)
= (1/12) * (i * (i + 1)) * ((i*(i + 1)) + (2*i + 1) + 1)
= (1/12) * (i * (i + 1)) * ((i2 + 3 * i + 2)
= (1/12) * (i * (i + 1)) * ((i + 1) * (i + 2))
= (1/12) * (i * (i + 1)2 * (i + 2))
Below is the implementation of this approach:
C++
// CPP Program to find the sum of sum of // squares of first n natural number#include <bits/stdc++.h>using namespace std;// Function to find sum of sum of square// of first n natural numberint findSum(int n){ return (n * (n + 1) * (n + 1) * (n + 2)) / 12;}// Driven Programint main(){ int n = 3; cout << findSum(n) << endl; return 0;} |
Java
// Java Program to find the sum of sum of // squares of first n natural numberclass GFG { // Function to find sum of sum of // square of first n natural number static int findSum(int n) { return (n * (n + 1) * (n + 1) * (n + 2)) / 12; } // Driver Program public static void main(String[] args) { int n = 3; System.out.println(findSum(n) ); }}// This code is contributed by Arnab Kundu |
Python3
# Python3 Program to find the sum # of sum of squares of first n # natural number# Function to find sum of sum of# square of first n natural numberdef findSum(n): return ((n * (n + 1) * (n + 1) * (n + 2)) / 12)# Driven Programn = 3print(int(findSum(n)))# This code is contributed by # Prasad Kshirsagar |
C#
// C# Program to find the sum of sum of // squares of first n natural numberusing System;class GFG { // Function to find sum of sum of // square of first n natural number static int findSum(int n) { return (n * (n + 1) * (n + 1) * (n + 2)) / 12; } // Driver Program static public void Main() { int n = 3; Console.WriteLine(findSum(n) ); }}// This code is contributed by Arnab Kundu |
PHP
<?php// PHP Program to find the sum of sum of // squares of first n natural number// Function to find sum of sum of square// of first n natural numberfunction findSum($n){ return ($n * ($n + 1) * ($n + 1) * ($n + 2)) / 12;} // Driver Code $n = 3; echo findSum($n) ;// This code is contributed by nitin mittal?> |
Javascript
<script>// js Program to find the sum of sum of// squares of first n natural number// Function to find sum of sum of square// of first n natural numberfunction findSum($n){ return (n * (n + 1) *(n + 1) * (n + 2)) / 12;} // Driver Code n = 3; document.write(findSum(n)) ;// This code is contributed by sravan kumar</script> |
Output:
20
Time Complexity : O(1)
Auxiliary Space : O(1)
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