# 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)

• Difficulty Level : Medium
• Last Updated : 30 Aug, 2022

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

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 sumint 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 Codeint 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 Codeif __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 seriesusing System; class GFG{ // Function to return required sumstatic 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 codepublic 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



## Javascript



Output:

40

Time Complexity: O(log(n))

Auxiliary Space: O(1)

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