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Program to find the sum of the series (1/a + 2/a^2 + 3/a^3 + … + n/a^n)
  • Last Updated : 06 Dec, 2018

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 i-th 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++

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// 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

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Java

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// 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));
    }
}

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Python 3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

0.6666667

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