Sum of the first N terms of the series 2, 6, 12, 20, 30….

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

Sn = 2 + 6 + 12 + 20 + 30 … upto n terms

Examples:

Input: N = 2
Output: 8
Explanation: 2 + 6 = 8

Input: N = 4 
Output: 40
Explanation: 2 + 6+ 12 + 20 = 40

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

Sn = 2 + 6 + 12 + 20 + 30......
Sn = (1+1^2) + (2+2^2) + (3+3^2) + (4+4^2) +......
Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^2 + 2^2 + 3^2 + 4^2 + ...upto n terms)

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

Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^2 + 2^2 + 3^2 + 4^2 + ...upto n terms)
Sn = n*(n + 1)/2 + n*(n + 1)*(2*n + 1)/6

Below is the implementation of the 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 + n * (n + 1) * (2 * n + 1) / 6;
}

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

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

    return 0;
}

Java

// Java program to find sum of first n terms

import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{

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

    return n * (n + 1) / 2 + n *
            (n + 1) * (2 * n + 1) / 6;
}

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

    // find the Sn
    System.out.print("Sum = " + calculateSum(n));

}
}

Python3

# Python program to find sum of
# first n terms

# Function to calculate the sum
def calculateSum(n):
    return (n * (n + 1) // 2 + n *
           (n + 1) * (2 * n + 1) // 6)

# 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
static int calculateSum(int n)
{

    return n * (n + 1) / 2 + n *
               (n + 1) * (2 * n + 1) / 6;
}

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

    // find the Sn
    Console.WriteLine("Sum = " +
                       calculateSum(n));
}
}

// This code is contributed by inder_verma

PHP

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

// Function to calculate the sum
function calculateSum($n)
{
    return $n * ($n + 1) / 2 + $n *
                ($n + 1) * (2 * $n + 1) / 6;
}

// Driver code

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

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

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

Javascript

<script>

// Javascript program to find sum of first n terms

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

    return n * (n + 1) / 2 + n *
            (n + 1) * (2 * n + 1) / 6;
}

// Driver code

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

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

// This code is contributed by Mayank Tyagi

</script>

Output:

Sum = 20

Time Complexity: O(1), we are using only constant-time operations.
Auxiliary Space: O(1)