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++

// 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;
}

Java

// 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 contributed by
// Smitha Dinesh Semwal

Python3

# 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.

C#

// 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 contributed by
// vt_m

PHP

<?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.
?>

Javascript

<script>
// Javascript implementation to find
// the sum of the given series

// function to find the sum
// of the given series

function sumOfTheSeries(n)
{
    let sum = 0;
    for (let i = 1; i <= n; i++)
    {

        // first term of each
        // i-th term
        let k = 2;
        for (let j = 1; j <= i; j++)
        {
            sum += k;

            // next term
            k += 2;
        }
    }

    // required sum
    return sum;
}

// Driver program to test above
    let n = 5;
    document.write("Sum = " + sumOfTheSeries(n));

// This code is contributed by gfgking.
</script>

Output: 

Sum = 70

Efficient Approach:
Let a_n 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++

// 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)
{
    // 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;
}

Java

// 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)
    {

        // 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 contributed by
//Smitha Dinesh Semwal

Python3

# Python3 implementation to find
# the sum of the given series

# function 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.

C#

// 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)
    {

        // 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 contributed by
// vt_m

PHP

<?php
// PHP implementation
// to find the sum
// of the given series

// function 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)
?>

Javascript

<script>

// JavaScript Program to find the
// sum of the given series

    // function to find the sum
    // of the given series
    function sumOfTheSeries(n)
    {

        // sum of 1st n natural numbers
        let sum_n = (n * (n + 1) / 2);

        // sum of squares of 1st n natural
        // numbers
        let sum_sq_n = (n * (n + 1) / 2) *
                        (2 * n + 1) / 3;

        // required sum
        return (sum_n + sum_sq_n);
    }


// Driver code

        let  n = 5;

        document.write("Sum = " +
                    sumOfTheSeries(n));

</script>

Output:

Sum = 70

Time complexity: O(1) as it is performing constant operations

Auxiliary space: O(1)