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

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

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/*CPP program to print the sum of series */
#include<bits/stdc++.h>
using namespace std;
  
/*f unction 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;
}

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Java

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// 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.

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?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.
?>

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Output :

6.26667

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