Sum of squares of first n natural numbers

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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.*/

Python3

 # 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.*/

PHP



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++

 // CPP Program to find sum  // of square of first n // natural numbers #include 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 si contributed by Nikita Tiwari.*/

Python3

 # 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.*/

PHP



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

 // 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 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; }

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)>

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)>

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.>

PHP



Output:

30

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