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
CPP
#include<bits/stdc++.h>
using namespace std;
double sumOfSeries( double a, double num)
{
double res = 0,prev=1;
for ( int i = 1; i <= num; i++)
{
prev *= (a/i);
res = res + prev;
}
return (res);
}
int main()
{
double n = 5, a=2;
cout << sumOfSeries(a,n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
public static void main (String[] args)
{
double n = 5 , a = 2 ;
System.out.println(sumOfSeries(a, n));
}
static double sumOfSeries( double a, double n)
{
double res = 0 , prev = 1 ;
for ( int i = 1 ; i <= n; i++)
{
prev *= (a / i);
res = res + prev;
}
return (res);
}
}
|
Python3
from __future__ import division
def sumOfSeries(a,num):
res = 0
prev = 1
for i in range ( 1 , n + 1 ):
prev * = (a / i)
res = res + prev
return res
n = 5
a = 2
print ( round (sumOfSeries(a,n), 4 ))
|
C#
using System;
class GFG
{
public static void Main ()
{
double n = 5, a = 2;
Console.WriteLine(sumOfSeries(a, n));
}
static float sumOfSeries( double a, double n)
{
double res = 0, prev = 1;
for ( int i = 1; i <= n; i++)
{
prev *= (a / i);
res = res + prev;
}
return ( float )(res);
}
}
|
PHP
Javascript
Output :
6.26667
Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.
Last Updated :
12 Jul, 2022
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