Program to find Sum of a Series a^1/1! + a^2/2! + a^3/3! + a^4/4! +.......+ a^n/n! - GeeksforGeeks

02 Nov, 2021
Given two values ‘a’ and ‘n’, find sum of series a^1/1! + a^2/2! + a^3/3! + a^4/4! +.......+ a^n/n!.Examples :  
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
 
A simple solution is to one by one compute values of individual terms and keep adding them to result.We can find solution with only one loop. The idea is to just use previous values and multiply by (a/i) where i is the no of term which we need to find. 
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
Illustration:  
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
Complexity:O(n)  
CPP
Java
Python3
C#
PHP
Javascript
/*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 seriesimport 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 coden = 5a = 2print(round(sumOfSeries(a,n),4)) # This Code is Contributed# by Azkia Anam.
// C# program to print the// sum of seriesusing 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 seriesfunction 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>
Output :  
6.26667
Time Complexity: O(n)
Auxiliary Space: O(1)
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