Dada una serie donde x, y y n toman valores enteros. La tarea es encontrar la suma hasta el término n de la serie dada.
Ejemplos:
Input: x = 2, y = 2, n = 2 Output: 40 Input: x = 2, y = 4, n = 2 Output: 92
Enfoque: La serie dada es:
.
Por lo tanto, nuestro problema se reduce a encontrar la suma de dos series de GP .
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP program to find the sum of series #include <bits/stdc++.h> 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
<?php // PHP program to find the // sum of series // Function to return required sum function 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; $n = 2; // function call to print sum echo sum($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 sum function 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 code let x = 2; let y = 2; let n = 2; // function call to print sum document.write(sum(x, y, n)); // This code is contributed // by bobby </script>
Producción:
40
Complejidad de tiempo: O(log(n))
Publicación traducida automáticamente
Artículo escrito por Sanjit_Prasad y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA