Sum of the series 2 + (2+4) + (2+4+6) + (2+4+6+8) + …… + (2+4+6+8+….+2n)

Given a positive integer n. The problem is to find the sum of the given series 2 + (2+4) + (2+4+6) + (2+4+6+8) + …… + (2+4+6+8+….+2n), where i-th term in the series is the sum of first i even natural numbers.

Examples:

Input : n = 2
Output : 8
(2) + (2+4) = 8

Input : n = 5
Output : 70
(2) + (2+4) + (2+4+6) + (2+4+6+8) + (2+4+6+8+10) = 70



Naive Approach: Using two loops get the sum of each i-th term and then add those sum to the final sum.

C++

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// C++ implementation to find the sum
// of the given series
#include <bits/stdc++.h>
   
using namespace std;
   
// function to find the sum
// of the given series
int sumOfTheSeries(int n)
{
    int sum = 0;
    for (int i = 1; i <= n; i++) {
   
        // first term of each i-th term
        int k = 2;
        for (int j = 1; j <= i; j++) {
            sum += k;
   
            // next term
            k += 2;
        }
    }
   
    // required sum
    return sum;
}
   
// Driver program to test above
int main()
{
    int n = 5;
    cout << "Sum = "
         << sumOfTheSeries(n);
    return 0;
}

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Java

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// Java implementation to find the
// sum of the given series
class GFG{
  
    // function to find the sum
    // of the given series
    static int sumOfTheSeries(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++) {
      
            // first term of each i-th term
            int k = 2;
            for (int j = 1; j <= i; j++) {
                sum += k;
      
                // next term
                k += 2;
            }
        }
      
        // required sum
        return sum;
    }
      
    // Driver program to test above
    public static void main(String[] args)
    {
        int n = 5;
  
        System.out.printf("Sum = %d",
                     sumOfTheSeries(n));
    }
}
  
// This code is contriubted by
// Smitha Dinesh Semwal

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Python3

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# Python3 implementation to find 
# the sum of the given series
  
# function to find the sum
# of the given series
def sumOfTheSeries(n):
  
    sum = 0
    for i in range(0, n + 1):
        # first term of each i-th
        # term
        k = 2
        for j in range(1, i + 1):
            sum = sum + k;
  
            # next term
            k = k + 2
      
    # required sum
    return sum;
  
# Driver program to test above
n = 5
ans = sumOfTheSeries(n);
print (ans)
  
# This code is contributed by saloni1297.

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C#

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// C# implementation to find the
// sum of the given series
using System;
  
class GFG{
  
    // function to find the sum
    // of the given series
    static int sumOfTheSeries(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++) {
      
            // first term of each i-th term
            int k = 2;
            for (int j = 1; j <= i; j++) {
                sum += k;
      
                // next term
                k += 2;
            }
        }
      
        // required sum
        return sum;
    }
      
    // Driver program to test above
    public static void Main()
    {
        int n = 5;
  
        Console.Write("Sum = "+
                    sumOfTheSeries(n));
    }
}
  
// This code is contriubted by
// vt_m

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PHP

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<?php
// PHP implementation to find
// the sum of the given series
  
// function to find the sum
// of the given series
  
function sumOfTheSeries($n)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
    {
  
        // first term of each 
        // i-th term
        $k = 2;
        for ($j = 1; $j <= $i; $j++) 
        {
            $sum += $k;
  
            // next term
            $k += 2;
        }
    }
  
    // required sum
    return $sum;
}
  
// Driver program to test above
    $n = 5;
    echo "Sum = ", sumOfTheSeries($n);
      
// This code is contributed by ajit.
?>

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Output:

Sum = 70

Efficient Approach:
Let an be the n-th term of the given series.

an = (2 + 4 + 6 + 8 +....+ 2n).
   = sum of first n even numbers.
   = n * (n + 1).
   = n2 + n.

Refer this post for the proof of above formula.

Now,

Refer this and this post for the proof of above formula.

C++

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// C++ implementation to find the sum
// of the given series
#include <bits/stdc++.h>
using namespace std;
   
// functionn to find the sum
// of the given series
int sumOfTheSeries(int n)
{
    // sum of 1st n natural numbers
    int sum_n = (n * (n + 1) / 2);
      
    // sum of squares of 1st n natural numbers
    int sum_sq_n = (n * (n + 1) / 2) *
                      (2 * n + 1) / 3;
                    
    // required sum
    return (sum_n + sum_sq_n);
}
   
// Driver program to test above
int main()
{
    int n = 5;
    cout << "Sum = "
         << sumOfTheSeries(n);
    return 0;
}

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Java

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// Java implementation to find the
// sum of the given series
class GFG{
      
    // functionn to find the sum
    // of the given series
    static int sumOfTheSeries(int n)
    {
          
        // sum of 1st n natural numbers
        int sum_n = (n * (n + 1) / 2);
          
        // sum of squares of 1st n natural
        // numbers
        int sum_sq_n = (n * (n + 1) / 2) *
                        (2 * n + 1) / 3;
                      
        // required sum
        return (sum_n + sum_sq_n);
    }
      
    // Driver program to test above
    public static void main(String[] args)
    {
        int n = 5;
          
        System.out.printf("Sum = %d",
                    sumOfTheSeries(n));
    }
}
  
// This code is contriubted by 
//Smitha Dinesh Semwal

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Python3

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# Python3 implementation to find 
# the sum of the given series
  
# functionn to find the sum
# of the given series
def sumOfTheSeries(n):
  
    # sum of 1st n natural numbers
    sum_n = int((n * (n + 1) / 2));
      
    # sum of squares of 1st n natural numbers
    sum_sq_n = int ((n * (n + 1) / 2) * (2 * n + 1) / 3)
                      
    # required sum
    return (sum_n + sum_sq_n);
  
# Driver program to test above
n = 5
ans = sumOfTheSeries(n)
print (ans)
  
# This code is contributed by saloni1297.

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C#

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// C# implementation to find the
// sum of the given series
using System;
  
class GFG{
      
    // functionn to find the sum
    // of the given series
    static int sumOfTheSeries(int n)
    {
          
        // sum of 1st n natural numbers
        int sum_n = (n * (n + 1) / 2);
          
        // sum of squares of 1st n 
        // natural numbers
        int sum_sq_n = (n * (n + 1) / 2) *
                        (2 * n + 1) / 3;
                      
        // required sum
        return (sum_n + sum_sq_n);
    }
      
    // Driver program to test above
    public static void Main()
    {
        int n = 5;
          
        Console.Write("Sum = "+
                    sumOfTheSeries(n));
    }
}
  
// This code is contriubted by 
// vt_m

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PHP

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<?php
// PHP implementation 
// to find the sum
// of the given series
  
// functionn to find the 
// sum of the given series
function sumOfTheSeries($n)
{
    // sum of 1st n
    // natural numbers
    $sum_n = ($n * ($n + 1) / 2);
      
    // sum of squares of
    // 1st n natural numbers
    $sum_sq_n = ($n * ($n + 1) / 2) *
                  (2 * $n + 1) / 3;
                      
    // required sum
    return ($sum_n +
            $sum_sq_n);
}
  
// Driver Code
$n = 5;
echo ("Sum = ".sumOfTheSeries($n));
  
// This code is contributed by 
// Manish Shaw(manishshaw1)
?>

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Output:

Sum = 70


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