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# 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
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```

## CPP

 `/*CPP program to print the sum of series */` `#include` `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

 `// 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.`

## Python3

 `# 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#

 `// 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

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## Javascript

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Output :

`6.26667`

Time Complexity: O(n)

Auxiliary Space: O(1), since no extra space has been taken.

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