Sum of the first N terms of the series 2,10, 30, 68,….

Given a number N, the task is to find the sum of first N terms of the below series:

Sn = 2 + 10 + 30 + 68 + … upto n terms

Examples:

Input: N = 2
Output: 12
2 + 10
= 12

Input: N = 4 
Output: 40
2 + 10 + 30 + 68
= 110

Approach: Let, the nth term be denoted by tn. 
This problem can easily be solved by splitting each term as follows : 

Sn = 2 + 10 + 30 + 68 + ......
Sn = (1+1^3) + (2+2^3) + (3+3^3) + (4+4^3) +......
Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^3 + 2^3 + 3^3 + 4^3 + ...upto n terms)

We observed that Sn can broken down into summation of two series. 
Hence, the sum of first n terms is given as follows: 

Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^3 + 2^3 + 3^3 + 4^3 + ...upto n terms)
Sn = n*(n + 1)/2 + (n*(n + 1)/2)^2

Below is the implementation of above approach:

C++

// C++ program to find sum of first n terms
#include <bits/stdc++.h>
using namespace std;

// Function to calculate the sum
int calculateSum(int n)
{

    return n * (n + 1) / 2
           + pow((n * (n + 1) / 2), 2);
}

// Driver code
int main()
{
    // number of terms to be
    // included in the sum
    int n = 3;

    // find the Sum
    cout << "Sum = " << calculateSum(n);

    return 0;
}

Java

// Java program to find sum of first n terms

public class GFG {

    // Function to calculate the sum
    static int calculateSum(int n)
    {

        return n * (n + 1) / 2
               + (int)Math.pow((n * (n + 1) / 2), 2);
    }

    // Driver code
    public static void main(String args[])
    {
        // number of terms to be
        // included in the sum
        int n = 3;

        // find the Sum
        System.out.println("Sum = "+ calculateSum(n));
    }
    // This Code is contributed by ANKITRAI1
}

Python3

# Python program to find sum
# of first n terms

# Function to calculate the sum
def calculateSum(n):
    return (n * (n + 1) // 2 +
        pow((n * (n + 1) // 2), 2))

# Driver code

# number of terms to be
# included in the sum
n = 3

# find the Sum
print("Sum = ", calculateSum(n))

# This code is contributed by
# Sanjit_Prasad

C#

// C# program to find sum of first n terms
using System;
class gfg
{
    // Function to calculate the sum
    public void calculateSum(int n)
    {
        double r = (n * (n + 1) / 2 +
                Math.Pow((n * (n + 1) / 2), 2));
        Console.WriteLine("Sum = " + r);
    }

    // Driver code
    public static int Main()
    {
        gfg g = new gfg();

        // number of terms to be
        // included in the sum
        int n = 3;

        // find the Sum
        g.calculateSum(n);
        Console.Read();
        return 0;
    }
}

PHP

<?php
// PHP program to find sum
// of first n terms

// Function to calculate the sum
function calculateSum($n)
{
    return $n * ($n + 1) / 2 +
      pow(($n * ($n + 1) / 2), 2);
}

// Driver code

// number of terms to be
// included in the sum
$n = 3;

// find the Sum
echo "Sum = " , calculateSum($n);

// This code is contributed
// by anuj_67
?>

Javascript

<script>

// Javascript program to find sum of first n terms

// Function to calculate the sum
function calculateSum(n)
{

    return n * (n + 1) / 2
        + Math.pow((n * (n + 1) / 2), 2);
}

// Driver code

    // number of terms to be
    // included in the sum
    let n = 3;

    // find the Sum
    document.write("Sum = " + calculateSum(n));


// This code is contributed by Mayank Tyagi

</script>

Output:

Sum = 42

Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.