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 <bits/stdc++.h>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 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
<?php// PHP program to find the // sum of series// Function to return required sumfunction sum($x, $y, $n){ //sum of first series $sum1 = (pow($x, 2) * (pow($x, 2 * $n) - 1)) / (pow($x, 2) - 1); // sum of second series $sum2 = ($x * $y * (pow($x, $n) * pow($y, $n) - 1)) / ($x * $y - 1); return $sum1 + $sum2;}// Driver code$x = 2;$y = 2;$n = 2;// function call to print sumecho sum($x, $y, $n);// This code is contributed // by Shashank_Sharma?> |
Javascript
<script>// java script program to find the // sum of series// Function to return required sumfunction sum(x, y, n){ //sum of first series sum1 = (Math.pow(x, 2) * (Math.pow(x, 2 * n) - 1)) / (Math.pow(x, 2) - 1); // sum of second series sum2 = (x * y * (Math.pow(x, n) * Math.pow(y, n) - 1)) / (x * y - 1); return sum1 + sum2;}// Driver codelet x = 2;let y = 2;let n = 2;// function call to print sumdocument.write(sum(x, y, n));// This code is contributed // by bobby</script> |
Output:
40
Time Complexity: O(log(n))
Auxiliary Space: O(1)
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 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
