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

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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)

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