# Find the sum of the series 2, 5, 13, 35, 97…

Given a series and a number n, the task is to find the sum of its first n terms. Below is the series:

2, 5, 13, 35, 97, …

Examples:

Input: N = 2
Output: 7
The sum of first 2 terms of Series is
2 + 5 = 7

Input: N = 4
Output: 55
The sum of first 4 terms of Series is
2 + 5 + 13 + 35 = 55


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: From this given series we find it is the sum of the Two GP series with common ratioes 2, 3 .

Sn = 2 + 5 + 13 + 35 + 97 … + upto nth term
Sn = (2^0 + 3^ 0) + (2^1 + 3^1) + (2^2 + 3^2) + (2^3 + 3^3)+ (2^4 + 3^4) …… + upto nth term
Sn = (2^0 + 2^1 + 2^2 + 2^3 + 2^4 … + upto nth term) + ( 3^0 + 3^1 + 3^2 + 3^3 …… + unto nth term )

Since, We know that the sum of n terms of the GP is given by the following formula: Below is the implementation of the above approach:

## C++

 // C++ program for finding the sum  // of first N terms of the series.  #include  using namespace std;     // CalculateSum function returns the final sum  int calculateSum(int n)  {      // r1 and r2 are common ratios      // of 1st and 2nd series      int r1 = 2, r2 = 3;         // a1 and a2 are common first terms      // of 1st and 2nd series      int a1 = 1, a2 = 1;         return a1 * (pow(r1, n) - 1) / (r1 - 1)             + a2 * (pow(r2, n) - 1) / (r2 - 1);  }     // Driver code  int main()  {      int n = 4;         // function calling for 4 terms      cout << "Sum = " << calculateSum(n)           << endl;         return 0;  }

## Java

 //Java program for finding the sum  //of first N terms of the series.     public class GFG {         //CalculateSum function returns the final sum      static int calculateSum(int n)      {       // r1 and r2 are common ratios       // of 1st and 2nd series       int r1 = 2, r2 = 3;          // a1 and a2 are common first terms       // of 1st and 2nd series       int a1 = 1, a2 = 1;          return (int)(a1 * (Math.pow(r1, n) - 1) / (r1 - 1)              + a2 * (Math.pow(r2, n) - 1) / (r2 - 1));      }         //Driver code      public static void main(String[] args) {                     int n = 4;              // function calling for 4 terms           System.out.println("Sum = " +calculateSum(n));      }  }

## Python 3

 # Python 3 program for finding the sum   # of first N terms of the series.      # from math import everything  from math import *    # CalculateSum function returns the final sum   def calculateSum(n) :         # r1 and r2 are common ratios       # of 1st and 2nd series       r1, r2 = 2, 3        # a1 and a2 are common first terms       # of 1st and 2nd series       a1, a2 = 1, 1        return (a1 * (pow(r1, n) - 1) // (r1 - 1)              + a2 * (pow(r2, n) - 1) // (r2 - 1))     # Driver Code  if __name__ == "__main__" :         n = 4        # function calling for 4 terms      print("SUM = ",calculateSum(n))        # This code is contributed by ANKITRAI1

## C#

 // C# program for finding the sum  // of first N terms of the series.  using System;     class GFG   {     // CalculateSum function   // returns the final sum  static int calculateSum(int n)  {  // r1 and r2 are common ratios  // of 1st and 2nd series  int r1 = 2, r2 = 3;     // a1 and a2 are common first   // terms of 1st and 2nd series  int a1 = 1, a2 = 1;     return (int)(a1 * (Math.Pow(r1, n) - 1) / (r1 - 1) +                a2 * (Math.Pow(r2, n) - 1) / (r2 - 1));  }     // Driver code  static public void Main ()  {      int n = 4;         // function calling for 4 terms      Console.Write("Sum = " +                     calculateSum(n));  }  }     // This code is contributed by Raj

## PHP

 

Output:

Sum = 55


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