# Find the sum of the series x(x+y) + x^2(x^2+y^2) +x^3(x^3+y^3)+ … + x^n(x^n+y^n)

Given a series where x, y and n take integral values. The task is to Find the sum till nth term of the given series.

Examples:

Input: x = 2, y = 2, n = 2
Output: 40

Input: x = 2, y = 4, n = 2
Output: 92


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

Approach: Given series is :

.

Thus our problem reduces to finding sum of two GP series.

Below is the implementation of above approach:

## C++

 // CPP program to find the sum of series  #include  using namespace std;     // Function to return required sum  int sum(int x, int y, int n)  {         // sum of first series      int sum1 = (pow(x, 2) * (pow(x, 2 * n) - 1))                 / (pow(x, 2) - 1);         // sum of second series      int sum2 = (x * y * (pow(x, n) * pow(y, n) - 1))                 / (x * y - 1);         return sum1 + sum2;  }     // Driver Code  int main()  {      int x = 2, y = 2, n = 2;         // function call to print sum      cout << sum(x, y, n);      return 0;  }

## Java

 // Java program to find the sum of series      public class GFG {             // Function to return required sum       static int sum(int x, int y, int n)       {                  // sum of first series           int sum1 = (int) (( Math.pow(x, 2) * (Math.pow(x, 2 * n) - 1))                      / (Math.pow(x, 2) - 1));                  // sum of second series           int sum2 = (int) ((x * y * (Math.pow(x, n) * Math.pow(y, n) - 1))                      / (x * y - 1));                  return sum1 + sum2;       }              // Driver code       public static void main (String args[]){          int x = 2, y = 2, n = 2;                      // function call to print sum           System.out.println(sum(x, y, n));       }     // This code is contributed by ANKITRAI1  }

## Python3

 # Python3 program to find the sum of series   # Function to return required sum      def sum(x,y,n):             # sum of first series       sum1 = ((x**2)*(x**(2*n)-1))//(x**2 - 1)             # sum of second series      sum2 = (x*y*(x**n*y**n-1))//(x*y-1)      return (sum1+sum2)         # Driver Code   if __name__=='__main__':      x = 2     y = 2     n = 2 # function call to print sum       print(sum(x, y, n))         # this code is contributed by sahilshelangia

## C#

 // C# program to find the sum of series   using System;     class GFG   {     // Function to return required sum   static int sum(int x, int y, int n)   {          // sum of first series       int sum1 = (int) ((Math.Pow(x, 2) *                         (Math.Pow(x, 2 * n) - 1)) /                        (Math.Pow(x, 2) - 1));          // sum of second series       int sum2 = (int) ((x * y * (Math.Pow(x, n) *                  Math.Pow(y, n) - 1)) / (x * y - 1));          return sum1 + sum2;   }      // Driver code   public static void Main ()  {      int x = 2, y = 2, n = 2;              // function call to print sum       Console.Write(sum(x, y, n));   }  }     // This code is contributed by ChitraNayal

## PHP

 

Output:

40


Time Complexity: O(log(n))

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