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>
long power(int a, int b)
{
    long i, p=1;
    for(i=1;i<=b;i++)
    {
        p=p*a;
    }
    return p;
}

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



Output:


sum: 701.000000




#include <stdio.h>
#include <stdlib.h>
#include <math.h>

long power(int a, int b)
{
    long i, p=1;
    for(i=1;i<=b;i++)
    {
        p=p*a;
    }
    return p;
}

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)/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: 186276539.000000
Process returned 0 (0x0) execution time : 1.742 s




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