Sum of squares of first n natural numbers (original) (raw)

Last Updated : 8 Aug, 2024

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

Input : 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++ `

// 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; }

Java

// 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./

Python

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.*/

C#

// 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.*/

JavaScript

PHP

i<=i <= i<=n; $i++) sum+=(sum += (sum+=(i * $i); return $sum; } // Driven Code $n = 4; echo(squaresum($n)); // This code is contributed by Ajit. ?>

`

**Output :

30

**Time Complexity: O(n)

**Auxiliary Space: O(1)

**Method 2: O(1)

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

For 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 * ? k2 + 3 * ? k + ? 1
n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * ? k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 3 * n = 3 * ? k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * ? k2
n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * ? k2
n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * ? k2
n * (n + 1) * (n + 2 - 3/2) = 3 * ? k2
n * (n + 1) * (2 * n + 1)/2 = 3 * ? k2
n * (n + 1) * (2 * n + 1)/6 = ? k2

Below is the implementation of this approach:

C++ `

// 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; }

Java

// 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 is contributed by Nikita Tiwari./

Python

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.

C#

// 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./

JavaScript

PHP

`

**Output :

30

**Time Complexity: O(1)

**Auxiliary Space: O(1), since no extra space has been taken

**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.

C++ `

// 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; }

Java

// 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)>

Python3

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)>

C#

// 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.>

JavaScript

PHP

`

**Output:

30

**Time complexity: O(1) since performing constant operations

**Space complexity: O(1) since using constant variables