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

• Last Updated : 31 May, 2022

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

Sn = 2 + 10 + 30 + 68 + â€¦ 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 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;     } }



## Javascript



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.

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