Skip to content
Related Articles

Related Articles

Sum of squares of first n natural numbers
  • Difficulty Level : Easy
  • Last Updated : 07 May, 2018

Given a positive integer N. The task is to find 12 + 22 + 32 + ….. + N2.

Examples :

Input : N = 4
Output : 30
12 + 22 + 32 + 42
= 1 + 4 + 9 + 16
= 30

Iput : N = 5
Output : 55

Method 1: O(N) The idea is to run a loop from 1 to n and for each i, 1 <= i <= n, find i2 to sum.

Below is the implementation of this approach

C++



filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP Program to find sum of square of first n natural numbers
#include <bits/stdc++.h>
using namespace std;
  
// Return the sum of square of first n natural numbers
int squaresum(int n)
{
    // Iterate i from 1 and n
    // finding square of i and add to sum.
    int sum = 0;
    for (int i = 1; i <= n; i++)
        sum += (i * i);
    return sum;
}
  
// Driven Program
int main()
{
    int n = 4;
    cout << squaresum(n) << endl;
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to find sum of 
// square of first n natural numbers
import java.io.*;
  
class GFG {
      
    // Return the sum of square of first n natural numbers
    static int squaresum(int n)
    {
        // Iterate i from 1 and n
        // finding square of i and add to sum.
        int sum = 0;
        for (int i = 1; i <= n; i++)
            sum += (i * i);
        return sum;
    }
       
    // Driven Program
    public static void main(String args[])throws IOException
    {
        int n = 4;
        System.out.println(squaresum(n));
    }
}
  
/*This code is contributed by Nikita Tiwari.*/

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 Program to
# find sum of square
# of first n natural 
# numbers
  
  
# Return the sum of
# square of first n
# natural numbers
def squaresum(n) :
  
    # Iterate i from 1 
    # and n finding 
    # square of i and
    # add to sum.
    sm = 0
    for i in range(1, n+1) :
        sm = sm + (i * i)
      
    return sm
  
# Driven Program
n = 4
print(squaresum(n))
  
# This code is contributed by Nikita Tiwari.*/

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# Program to find sum of
// square of first n natural numbers
using System;
  
class GFG {
  
    // Return the sum of square of first
    // n natural numbers
    static int squaresum(int n)
    {
          
        // Iterate i from 1 and n
        // finding square of i and add to sum.
        int sum = 0;
          
        for (int i = 1; i <= n; i++)
            sum += (i * i);
              
        return sum;
    }
  
    // Driven Program
    public static void Main()
    {
        int n = 4;
          
        Console.WriteLine(squaresum(n));
    }
}
  
/* This code is contributed by vt_m.*/

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP Program to find sum of 
// square of first n natural numbers
  
// Return the sum of square of
// first n natural numbers
function squaresum($n)
{
    // Iterate i from 1 and n
    // finding square of i and 
    // add to sum.
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
        $sum += ($i * $i);
    return $sum;
}
  
// Driven Code
$n = 4;
echo(squaresum($n));
  
// This code is contributed by Ajit.
?>

chevron_right



Output :

30

Method 2: O(1)

Sum of squares of first N natural numbers = (N*(N+1)*(2*N+1))/6

Foe example
For N=4, Sum = ( 4 * ( 4 + 1 ) * ( 2 * 4 + 1 ) ) / 6
= 180 / 6
= 30
For N=5, Sum = ( 5 * ( 5 + 1 ) * ( 2 * 5 + 1 ) ) / 6
= 55

Proof:

We know,
(k + 1)3 = k3 + 3 * k2 + 3 * k + 1
We can write the above identity for k from 1 to n:
23 = 13 + 3 * 12 + 3 * 1 + 1 ......... (1)
33 = 23 + 3 * 22 + 3 * 2 + 1 ......... (2)
43 = 33 + 3 * 32 + 3 * 3 + 1 ......... (3)
53 = 43 + 3 * 42 + 3 * 4 + 1 ......... (4)
...
n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ......... (n - 1)
(n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ......... (n)

Putting equation (n - 1) in equation n,
(n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1
         = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1

By putting all equation, we get
(n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1
n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2
n * (n + 1) * (2 * n + 1)/2  = 3 * Σ k2
n * (n + 1) * (2 * n + 1)/6  = Σ k2

Below is the implementation of this approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP Program to find sum 
// of square of first n
// natural numbers
#include <bits/stdc++.h>
using namespace std;
  
// Return the sum of square of
// first n natural numbers
int squaresum(int n)
{
    return (n * (n + 1) * (2 * n + 1)) / 6;
}
  
// Driven Program
int main()
{
    int n = 4;
    cout << squaresum(n) << endl;
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to find sum 
// of square of first n
// natural numbers
import java.io.*;
  
class GFG {
      
    // Return the sum of square 
    // of first n natural numbers
    static int squaresum(int n)
    {
        return (n * (n + 1) * (2 * n + 1)) / 6;
    }
      
    // Driven Program
    public static void main(String args[])
                            throws IOException
    {
        int n = 4;
        System.out.println(squaresum(n));
    }
}
  
  
/*This code si contributed by Nikita Tiwari.*/

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 Program to
# find sum of square 
# of first n natural 
# numbers
  
# Return the sum of 
# square of first n
# natural numbers
def squaresum(n) :
    return (n * (n + 1) * (2 * n + 1)) // 6
  
# Driven Program
n = 4
print(squaresum(n))
  
#This code is contributed by Nikita Tiwari.                                                               

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# Program to find sum
// of square of first n
// natural numbers
using System;
  
class GFG {
  
    // Return the sum of square
    // of first n natural numbers
    static int squaresum(int n)
    {
        return (n * (n + 1) * (2 * n + 1)) / 6;
    }
  
    // Driven Program
    public static void Main()
  
    {
        int n = 4;
          
        Console.WriteLine(squaresum(n));
    }
}
  
/*This code is contributed by vt_m.*/

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP Program to find sum 
// of square of first n
// natural numbers
  
// Return the sum of square of
// first n natural numbers
function squaresum($n)
{
    return ($n * ($n + 1) * 
           (2 * $n + 1)) / 6;
}
  
// Driven Code
$n = 4;
echo(squaresum($n));
  
// This code is contributed by Ajit.
?>

chevron_right



Output :

30

Avoiding early overflow:
For large n, the value of (n * (n + 1) * (2 * n + 1)) would overflow. We can avoid overflow up to some extent using the fact that n*(n+1) must be divisible by 2.

CPP

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP Program to find sum of square of first
// n natural numbers. This program avoids
// overflow upto some extent for large value
// of n.
#include <bits/stdc++.h>
using namespace std;
  
// Return the sum of square of first n natural
// numbers
int squaresum(int n)
{
    return (n * (n + 1) / 2) * (2 * n + 1) / 3;
}
  
// Driven Program
int main()
{
    int n = 4;
    cout << squaresum(n) << endl;
    return 0;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python Program to find sum of square of first
# n natural numbers. This program avoids
# overflow upto some extent for large value
# of n.y
  
def squaresum(n):
    return (n * (n + 1) / 2) * (2 * n + 1) / 3
  
# main()
n = 4
print(squaresum(n));
  
# Code Contributed by Mohit Gupta_OMG <(0_o)>

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to find sum of square of first
// n natural numbers. This program avoids
// overflow upto some extent for large value
// of n.
  
import java.io.*;
import java.util.*; 
  
class GFG
{
    // Return the sum of square of first n natural
    // numbers
public static int squaresum(int n)
{
    return (n * (n + 1) / 2) * (2 * n + 1) / 3;
}
  
    public static void main (String[] args)
    {
        int n = 4;
    System.out.println(squaresum(n));
    }
}
  
// Code Contributed by Mohit Gupta_OMG <(0_o)>

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# Program to find sum of square of first
// n natural numbers. This program avoids
// overflow upto some extent for large value
// of n.
  
using System;
  
class GFG {
      
    // Return the sum of square of
    // first n natural numbers
    public static int squaresum(int n)
    {
        return (n * (n + 1) / 2) * (2 * n + 1) / 3;
    }
  
    // Driver Code
    public static void Main()
    {
        int n = 4;
          
        Console.WriteLine(squaresum(n));
    }
}
  
// This Code is Contributed by vt_m.>

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP Program to find 
// sum of square of first
// n natural numbers. 
// This program avoids
// overflow upto some 
// extent for large value
// of n.
  
// Return the sum of square
// of first n natural numbers
function squaresum($n)
{
    return ($n * ($n + 1) / 2) * 
           (2 * $n + 1) / 3;
}
  
    // Driver Code
    $n = 4;
    echo squaresum($n) ;
      
// This code is contributed by vt_m.
?>

chevron_right



Output:

30

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :