Program to find the sum of the series (1/a + 2/a^2 + 3/a^3 + … + n/a^n)
Given two integers a and n. The task is to find the sum of the series 1/a + 2/a2 + 3/a3 + … + n/an.
Examples:
Input: a = 3, n = 3
Output: 0.6666667
The series is 1/3 + 1/9 + 1/27 which is
equal to 0.6666667
Input: a = 5, n = 4
Output: 0.31039998
Approach: Run a loop from 1 to n and get the term of the series by calculating term = (i / ai). Sum all the generated terms which is the final answer.
Below is the implementation of the above approach:
C++
// C++ program to find the sum of // the given series #include <stdio.h> #include <math.h> #include <iostream> using namespace std; // Function to return the // sum of the series float getSum( int a, int n) { // variable to store the answer float sum = 0; for ( int i = 1; i <= n; ++i) { // Math.pow(x, y) returns x^y sum += (i / pow (a, i)); } return sum; } // Driver code int main() { int a = 3, n = 3; // Print the sum of the series cout << (getSum(a, n)); return 0; } // This code is contributed // by Sach_Code |
Java
// Java program to find the sum of the given series import java.util.Scanner; public class HelloWorld { // Function to return the sum of the series public static float getSum( int a, int n) { // variable to store the answer float sum = 0 ; for ( int i = 1 ; i <= n; ++i) { // Math.pow(x, y) returns x^y sum += (i / Math.pow(a, i)); } return sum; } // Driver code public static void main(String[] args) { int a = 3 , n = 3 ; // Print the sum of the series System.out.println(getSum(a, n)); } } |
Python 3
# Python 3 program to find the sum of # the given series import math # Function to return the # sum of the series def getSum(a, n): # variable to store the answer sum = 0 ; for i in range ( 1 , n + 1 ): # Math.pow(x, y) returns x^y sum + = (i / math. pow (a, i)); return sum ; # Driver code a = 3 ; n = 3 ; # Print the sum of the series print (getSum(a, n)); # This code is contributed # by Akanksha Rai |
C#
// C# program to find the sum // of the given series using System; class GFG { // Function to return the sum // of the series public static double getSum( int a, int n) { // variable to store the answer double sum = 0; for ( int i = 1; i <= n; ++i) { // Math.pow(x, y) returns x^y sum += (i / Math.Pow(a, i)); } return sum; } // Driver code static public void Main () { int a = 3, n = 3; // Print the sum of the series Console.WriteLine(getSum(a, n)); } } // This code is contributed by jit_t |
PHP
<?php // PHP program to find the // sum of the given series // Function to return the // sum of the series function getSum( $a , $n ) { // variable to store the answer $sum = 0; for ( $i = 1; $i <= $n ; ++ $i ) { // Math.pow(x, y) returns x^y $sum += ( $i / pow( $a , $i )); } return $sum ; } // Driver code $a = 3; $n = 3; // Print the sum of the series echo (getSum( $a , $n )); // This code is contributed by akt_mit ?> |
Javascript
<script> // Javascript program to find the sum of the given series // Function to return the sum of the series function getSum( a, n) { // variable to store the answer let sum = 0; for (let i = 1; i <= n; ++i) { // Math.pow(x, y) returns x^y sum += (i / Math.pow(a, i)); } return sum; } // Driver code let a = 3, n = 3; // Print the sum of the series document.write(getSum(a, n).toFixed(7)); // This code contributed by Princi Singh </script> |
Output:
0.6666667
Time Complexity: O(nlogn)
Auxiliary Space: O(1)
Method: Finding the sum of series without using pow function
C++
#include <iostream> #include<math.h> using namespace std; int main() { // sum of series int n = 3, a = 3; double s = 0; // iterating for loop n times for ( int i = 1; i < n + 1; i++) { // finding sum s = s + (i / ( pow (a, i))); } // printing the result cout << s; return 0; } // This code is contributed by ksrikanth0498. |
Java
/*package whatever //do not write package name here */ // Java code to print // sum of series import java.io.*; class GFG { public static void main (String[] args) { int n = 3 ,a = 3 ,s = 0 ; //iterating for loop n times for ( int i = 1 ; i <= n; i++) { //finding sum s = s + (i/(a**i)) } //printing the result System.out.println(s); } } // This code is contributed by Atul_kumar_Shrivastava |
Python3
# Python code to print # sum of series n = 3 ; a = 3 ; s = 0 # iterating for loop n times for i in range ( 1 , n + 1 ): # finding sum s = s + (i / (a * * i)) # printing the result print (s) # this code is contributed by Gangarajula Laxmi |
C#
using System; public class GFG { static public void Main () { // sum of series int n = 3, a = 3, s = 0; // iterating for loop n times for ( int i = 1; i < n + 1; i++) { // finding sum s = s + (i / (Math.Pow(a, i))); } // printing the result Console.WriteLine(s); } // this code is contributed by Gangarajula Laxmi |
Javascript
<script> // JavaScript code for the above approach //sum of series let n = 3, a = 3, s = 0; // iterating for loop n times for (let i = 1; i < n + 1; i++) { // finding sum s = s + (i / (Math.pow(a, i))) } // printing the result document.write(s); // This code is contributed by Potta Lokesh </script> |
Output
0.6666666666666667
Time complexity: O(nlogn) since a single using for loop and logn for inbuilt pow() function.
Auxiliary Space: O(1)
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