¡Programa para encontrar la suma de una serie a^1/1! + a^2/2! + a^3/3! + a^4/4! +…….+ un^n/n!

Dados dos valores ‘a’ y ‘n’, encuentre la suma de la serie a^1/1. + a^2/2! + a^3/3! + a^4/4! +…….+ un^n/n!.
Ejemplos: 
 

Input : a = 2 and n = 5
Output : 6.26667
We get result by adding 
2^1/1! + 2^2/2! + 2^3/3! + 2^4/4! +
2^5/5! 
= 2/1 + 4/2 +  8/6 + 16/24 + 32/120
=  6.26667

Una solución simple es calcular uno por uno los valores de los términos individuales y seguir agregándolos al resultado.
Podemos encontrar la solución con un solo bucle. La idea es simplemente usar los valores anteriores y multiplicar por (a/i) donde i es el número del término que necesitamos encontrar.
 

for finding 1st term:-  a/1
for finding 2nd term:-  (1st term) * a/2
for finding 3rd term:-  (2nd term) * a/3
.
.
.
for finding nth term:-  ((n-1)th term) * a/n

Ilustración: 
 

Input: a = 2 and n = 5
By multiplying Each term by 2/i
  1st term :-              2/1 = 2
  2nd term :- (1st term) * 2/2 =(2)*1 = 2
  3rd term :- (2nd term) * 2/3 = 4/3
  4th term :- (3rd term) * 2/4 = 2/3
  5th term :- (4th term) * 2/5 = 4/15
=> 2 + 2 + 4/3 + 2/3 + 4/15
Output: sum = 6.26667

/*CPP program to print the sum of series */
#include<bits/stdc++.h>
using namespace std;
 
/*function to calculate sum of given series*/
double sumOfSeries(double a,double num)
{
    double res = 0,prev=1;
    for (int i = 1; i <= num; i++)
    {
        /*multiply (a/i) to previous term*/
        prev *= (a/i);
 
        /*store result in res*/
        res = res + prev;
    }
    return(res);
}
 
/* Driver Function */
int main()
{
    double n = 5, a=2;
    cout << sumOfSeries(a,n);
    return 0;
}

// Java program to print the
// sum of series
import java.io.*;
 
class GFG
{
    public static void main (String[] args)
    {
        double n = 5, a = 2;
        System.out.println(sumOfSeries(a, n));
    }
     
    // function to calculate sum of given series
    static double sumOfSeries(double a,double n)
    {
        double res = 0, prev = 1;
        for (int i = 1; i <= n; i++)
            {
                // multiply (a/i) to previous term
                prev *= (a / i);
                     
                // store result in res
                res = res + prev;
            }
        return(res);
    }
}
 
// This code is Contributed by Azkia Anam.

# Python program to print
# the sum of series.
# function to calculate
# sum of given series.
 
from __future__ import division
 
def sumOfSeries(a,num):
    res = 0
    prev=1
    for i in range(1, n+1):
 
        # multiply (a/i) to
        # previous term
        prev *= (a/i)
 
        # store result in res
        res = res + prev
    return res
 
# Driver code
n = 5
a = 2
print(round(sumOfSeries(a,n),4))
 
# This Code is Contributed
# by Azkia Anam.

// C# program to print the
// sum of series
using System;
 
class GFG
{
    public static void Main ()
    {
        double n = 5, a = 2;
        Console.WriteLine(sumOfSeries(a, n));
    }
     
    // Function to calculate sum of given series
    static float sumOfSeries(double a, double n)
    {
        double res = 0, prev = 1;
        for (int i = 1; i <= n; i++)
            {
                // multiply (a/i) to previous term
                prev *= (a / i);
                     
                // store result in res
                res = res + prev;
            }
        return(float)(res);
    }
}
 
// This code is Contributed by vt_m.

<?php
// PHP program to print
// the sum of series
 
// Function to calculate
// sum of given series
function sumOfSeries($a, $num)
{
    $res = 0; $prev = 1;
    for ($i = 1; $i <= $num; $i++)
    {
        // multiply (a/i) to
        // previous term
        $prev *= ($a / $i);
 
        // store result in res
        $res = $res + $prev;
    }
    return ($res);
}
 
// Driver Code
$n = 5; $a = 2;
echo(sumOfSeries($a, $n));
 
// This code is contributed by Ajit.
?>

<script>
/* JavaScript program to print the sum of series */
 
/*function to calculate sum of given series*/
function sumOfSeries(a, num)
{
    let res = 0, prev = 1;
    for (let i = 1; i <= num; i++)
    {
     
        /*multiply (a/i) to previous term*/
        prev *= (a/i);
 
        /*store result in res*/
        res = res + prev;
    }
    return(res);
}
 
/* Driver Function */
    let n = 5, a=2;
    document.write(sumOfSeries(a,n));
 
// This code is contributed by Surbhi Tyagi.
</script>

Producción : 
 

6.26667

Complejidad de tiempo: O(n)

Espacio Auxiliar: O(1), ya que no se ha ocupado ningún espacio extra.

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