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!.
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
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
This article is contributed by R_Raj. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.