C program to Calculate [(1^1)/1!] + [(2^2)/2!] + [(3^3)/3!] + [(4^4)/4!] + [(5^5)/5!] + ... + [(n^n)/n!] series

In c language you can print any number series. Here i will show you how to print number series in c language with explanation. In case of print number series you need to focus on common difference between two numbers..

C program to Calculate [(1^1)/1!] + [(2^2)/2!] + [(3^3)/3!] + [(4^4)/4!] + [(5^5)/5!] + ... + [(n^n)/n!] series




#include<stdio.h>
#include<math.h>
double power(int a, int b)
{
    long i, p=1;
    for(i=1;i<=b;i++)
    {
        p=p*a;
    }
    return p;
}

double fact(int n)
{
    long i, f=1;
    for(i=1;i<=n;i++)
    {
        f=f*i;
    }
    return f;
}

int main()
{
    long i,n;
    double sum=0;
    n=5;
    for(i=1;i<=n;i++)
    {
        sum=sum+power(i,i)/fact(i);
    }
    printf("Sum: %lf",sum);
    return 0;
}




Output:


Sum: 44.208333




#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double power(int a, int b)
{
    long i, p=1;
    for(i=1;i<=b;i++)
    {
        p=p*a;
    }
    return p;
}

double fact(int n)
{
    long i, f=1;
    for(i=1;i<=n;i++)
    {
        f=f*i;
    }
    return f;
}

int main()
{
    long i,n;
    double sum=0;
    printf("\n Enter n value [(1^1)/1!] + [(2^2)/2!] + [(3^3)/3!] + [(4^4)/4!] + [(5^5)/5!] + ... + [(n^n)/n!]:");
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        sum=sum+power(i,i)/fact(i);
    }
    printf("Sum: %lf",sum);
    return 0;
}




Output:

Enter n value [(1^1)/1!] + [(2^2)/2!] + [(3^3)/3!] + [(4^4)/4!] + [(5^5)/5!] +... + [(n^n)/n!] : 10
Sum: 2144.714544
Process returned 0 (0x0) execution time : 1.886 s




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