# 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

Approach: From this given series we find it is the sum of the Two GP series with common ratios 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

 

## Javascript

 

Output

Sum = 55


Time Complexity: O(log n)
Auxiliary Space: O(1), As constant extra space is used.

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next