Programa para encontrar la suma de la serie (1/a + 2/a^2 + 3/a^3 + … + n/a^n)

Dados dos enteros  y  . La tarea es encontrar la suma de la serie 1/a + 2/a ² + 3/a ³ + … + n/a ⁿ .
Ejemplos: 
 

Entrada: a = 3, n = 3 
Salida: 0,6666667 
La serie es 1/3 + 1/9 + 1/27 que es 
igual a 0,6666667
Entrada: a = 5, n = 4 
Salida: 0,31039998 
 

Enfoque: ejecute un ciclo de 1 a n y obtenga el  término de la serie calculando término = (i / a ⁱ ) . Sume todos los términos generados, que es la respuesta final.
A continuación se muestra la implementación del enfoque anterior:
 

// 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 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 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# 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 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
?>

<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>

0.6666667

Complejidad de tiempo: O (nlogn)

Espacio Auxiliar: O(1)

Método: encontrar la suma de series sin usar la función pow 

# 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

<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>

0.6666666666666667

Complejidad de tiempo : O (nlogn) desde un solo uso para loop y logn para la función pow() incorporada.

Espacio Auxiliar: O(1)