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Program to find the sum of a Series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + … + (n*n)

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  • Difficulty Level : Medium
  • Last Updated : 10 Nov, 2022
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You have been given a series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + … + (n*n), find out the sum of the series till nth term. 
Examples : 
 

Input : n = 3
Output : 14
Explanation : (1*1) + (2*2) + (3*3) 

Input : n = 5
Output : 55
Explanation : (1*1) + (2*2) + (3*3) + (4*4) + (5*5)

 

 

C++




// CPP program to calculate the following series
#include<iostream>
using namespace std;
 
// Function to calculate the following series
int Series(int n)
{
    int i;
    int sums = 0;
    for (i = 1; i <= n; i++)
        sums += (i * i);
    return sums;
}
 
// Driver Code
int main()
{
    int n = 3;
    int res = Series(n);
    cout<<res<<endl;
}

C




// C program to calculate the following series
#include <stdio.h>
 
// Function to calculate the following series
int Series(int n)
{
    int i;
    int sums = 0;
    for (i = 1; i <= n; i++)
        sums += (i * i);
    return sums;
}
 
// Driver Code
int main()
{
    int n = 3;
    int res = Series(n);
    printf("%d", res);
}

Java




// Java program to calculate the following series
import java.io.*;
class GFG {
 
    // Function to calculate the following series
    static int Series(int n)
    {
        int i;
        int sums = 0;
        for (i = 1; i <= n; i++)
            sums += (i * i);
        return sums;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 3;
        int res = Series(n);
        System.out.println(res);
    }
}

Python




# Python program to calculate the following series
def Series(n):
    sums = 0
    for i in range(1, n + 1):
        sums += (i * i);
    return sums
 
# Driver Code
n = 3
res = Series(n)
print(res)

C#




// C# program to calculate the following series
using System;
class GFG {
 
    // Function to calculate the following series
    static int Series(int n)
    {
        int i;
        int sums = 0;
        for (i = 1; i <= n; i++)
            sums += (i * i);
        return sums;
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 3;
        int res = Series(n);
        Console.Write(res);
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to calculate
// the following series
 
// Function to calculate the
// following series
function Series($n)
{
    $i;
    $sums = 0;
    for ($i = 1; $i <= $n; $i++)
        $sums += ($i * $i);
    return $sums;
}
 
// Driver Code
$n = 3;
$res = Series($n);
echo($res);
 
// This code is contributed by Ajit.
?>

Javascript




<script>
 
// Javascript program to calculate the following series
 
    // Function to calculate the following series
    function Series( n) {
        let i;
        let sums = 0;
        for (i = 1; i <= n; i++)
            sums += (i * i);
        return sums;
    }
 
    // Driver Code
      
        let n = 3;
        let res = Series(n);
        document.write(res);
 
// This code contributed by Princi Singh
 
</script>

Output

14

Time Complexity: O(n)

Auxiliary Space: O(1)

Efficient Approach:

As we know the sum of squares of the first N natural numbers = (N*(N+1)*(2*N+1))/6.

Below is the implementation of this approach: 

C++




// CPP program to calculate the following series
#include<iostream>
using namespace std;
 
// Function to calculate the following series
int Series(int n)
{
    return (n * (n + 1) * (2 * n + 1)) / 6;
}
 
// Driver Code
int main()
{
    int n = 3;
    int res = Series(n);
    cout<<res<<endl;
}

Java




// Java program to calculate the following series
import java.io.*;
class GFG {
 
  // Function to calculate the following series
  static int Series(int n)
  {
    return (n * (n + 1) * (2 * n + 1)) / 6;
  }
 
  // Driver function
  public static void main (String[] args) {
    int n = 3;
    int res = Series(n);
    System.out.println(res);
 
  }
}
 
// This code is contributed by Aman Kumar.

Python3




# Python program to calculate the following series
 
# Function to calculate the following series
def Series(n):
    return (n * (n+1) * (2 * n + 1)) / 6
 
n = 3
res = Series(n)
print(int(res))
 
# This code is contributed by lokeshmvs21.

C#




using System;
 
public class GFG {
  static int Series(int n)
  {
    return (n * (n + 1) * (2 * n + 1)) / 6;
  }
 
  static public void Main()
  {
 
    int n = 3;
    Console.Write(Series(n));
  }
}
 
// This code is contributed by garg28harsh.

Javascript




// Javascript program to calculate the following series
 
    // Function to calculate the following series
    function Series(n)
{
    return (n * (n + 1) * (2 * n + 1)) / 6;
}
 
// Driver Code
let n = 3;
console.log(Series(n));
 
// This code is contributed by garg28harsh.

Output

14

Time Complexity: O(1)

Auxiliary Space: O(1)
Please refer below post for O(1) solution. 
Sum of squares of first n natural numbers
 


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