Q:

# C program to find the sum of series 1^2/1! + 2^2/2! + 3^2/3! + 4^2/4! + ... n^2/n!

Given the value of n and we have to find the sum of series 1^2/1! + 2^2/2! + 3^2/3! + 4^2/4! + ... n^2/n! Using C program.

## Program to find the sum of series 1^2/1! + 2^2/2! + 3^2/3! + 4^2/4! + ... n^2/n!

``````/*
C program to find sum of following series
* 1^2/1! + 2^2/2! + 3^2/3! + 4^2/4! + ... n^2/n!
*/

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

/* function : factorial, to find factorial of a given number*/
long int factorial(int n)
{
int i;
long int fact=1;

if(n==1) return fact;

for(i=n;i>=1;i--)
fact= fact * i;

return fact;
}

// main function
int main()
{
int i,N,x;
float sum;

printf("Enter total number of terms: ");
scanf("%d",&N);

/*set sum by 0*/
sum=0.0f;

/*calculate sum of the series*/
for(i=1;i<=N;i++)
{
sum = sum + ( (float)(pow(N,2)) / (float)(factorial(N)));
}

/*print the sum*/
printf("Sum of the series is: %f\n",sum);

return 0;
}``````

Output

```Run(1)
Enter total number of terms: 3
Sum of the series is: 4.500000

Run(2)
Enter total number of terms: 4
Sum of the series is: 2.666667```