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

• Last Updated : 22 Jun, 2022

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 â€¦ 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 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



## Javascript



Output:

Sum = 20

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

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