Problem statement:- Program to find the sum of series 1^1/1!+2^2/2!+3^3/3!...+n^n/n!
Output Data:-Sum
Additional Data:-i, fact
Input/Output:
Input/Output:
Input/Output:
Input/Output:
Data requirement:-
Input Data:- n
Output Data:-Sum
Additional Data:-i, fact
Program in C
Here is the source code of the C Program to find the sum of series 1^1/1!+2^2/2!+ 3^3/3!...+n^n/n!.
Code:
#include<stdio.h>
#include<math.h>
int main()
{
int n,i,fact=1;
double sum=0.0;
printf("Enter the range of number:");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
fact*=i;
sum+=pow(i,i)/fact;
}
printf("The sum of the series = %0.2lf",sum);
}
#include<math.h>
int main()
{
int n,i,fact=1;
double sum=0.0;
printf("Enter the range of number:");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
fact*=i;
sum+=pow(i,i)/fact;
}
printf("The sum of the series = %0.2lf",sum);
}
Input/Output:
Enter the range of number:5
The sum of the series = 44.21
The sum of the series = 44.21
Program in C++
Here is the source code of the C++ Program to find the sum of series 1^1/1!+2^2/2!+ 3^3/3!...+n^n/n!.
Code:
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int n,i,fact;
double sum=0.0;
cout<<"Enter the range of number:";
cin>>n;
for(i=1;i<=n;i++)
{
fact*=i;
sum+=pow(i,i)/fact;
}
cout<<"The sum of the series = "<<sum;
}
#include<cmath>
using namespace std;
int main()
{
int n,i,fact;
double sum=0.0;
cout<<"Enter the range of number:";
cin>>n;
for(i=1;i<=n;i++)
{
fact*=i;
sum+=pow(i,i)/fact;
}
cout<<"The sum of the series = "<<sum;
}
Input/Output:
Enter the range of number:7
The sum of the series = 272.41
Program in Java
The sum of the series = 272.41
Program in Java
Here is the source code of the Java Program to find the sum of series 1^1/1!+2^2/2!+ 3^3/3!...+n^n/n!.
Code:
import java.util.Scanner;
public class p20 {
public static void main(String[] args) {
Scanner cs=new Scanner(System.in);
int n,i,fact=1;
double sum=0.0;
System.out.println("Enter the range of number:");
n=cs.nextInt();
for(i=1;i<=n;i++)
{
fact*=i;
sum+=Math.pow(i,i)/fact;
}
System.out.println("The sum of the series = "+sum);
cs.close();
}
}
public class p20 {
public static void main(String[] args) {
Scanner cs=new Scanner(System.in);
int n,i,fact=1;
double sum=0.0;
System.out.println("Enter the range of number:");
n=cs.nextInt();
for(i=1;i<=n;i++)
{
fact*=i;
sum+=Math.pow(i,i)/fact;
}
System.out.println("The sum of the series = "+sum);
cs.close();
}
}
Input/Output:
Enter the range of number:
9
The sum of the series = 1756.138318452381
Program in Python
9
The sum of the series = 1756.138318452381
Program in Python
Here is the source code of the Python Program to find the sum of series 1^1/1!+2^2/2!+ 3^3/3!...+n^n/n!.
import math
print("Enter the range of number:")
n=int(input())
sum=0.0
fact=1
for i in range(1,n+1):
fact*=i
sum += pow(i, i) / fact
print("The sum of the series = ",sum)
print("Enter the range of number:")
n=int(input())
sum=0.0
fact=1
for i in range(1,n+1):
fact*=i
sum += pow(i, i) / fact
print("The sum of the series = ",sum)
Input/Output:
Enter the range of number:
20
The sum of the series = 69257915.90725157
20
The sum of the series = 69257915.90725157
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