C Exercises: Calculate the sum of the series [ 1-X^2/2!+X^4/4!- .........]

C For Loop: Exercise-18 with Solution

Write a program in C to find the sum of the series [ 1-X^2/2!+X^4/4!- .........].

This C program calculates the sum of a series where the terms alternate between positive and negative, involving powers of X and factorial denominators: 1-X^2/2!+X^4/4!-.. . The program will take an input value for X and a specified number of terms, then compute the sum of the series using a loop to iterate through each term, applying the appropriate sign and factorial calculation.

Sample Solution:

C Code:

#include <stdio.h>  // Include the standard input/output header file.

void main()
{
    float x, sum, t, d;  // Declare variables to store input and intermediate values.
    int i, n;  // Declare variables for loop control and input.

    // Prompt the user to input the value of 'x'.
    printf("Input the Value of x :");
    scanf("%f", &x);  // Read the value of 'x' from the user.

    // Prompt the user to input the number of terms.
    printf("Input the number of terms : ");
    scanf("%d", &n);  // Read the value of 'n' from the user.

    sum = 1;  // Initialize 'sum' to 1, as the first term is always 1.
    t = 1;    // Initialize 't' to 1 for the first term.

    // Loop to calculate the sum of the series.
    for (i = 1; i < n; i++)
    {
        d = (2 * i) * (2 * i - 1);  // Calculate the denominator for the term.
        t = -t * x * x / d;        // Calculate the term value.
        sum = sum + t;             // Add the term to the sum.
    }

    // Print the final result along with the input values.
    printf("\nThe sum = %f\nNumber of terms = %d\nValue of x = %f\n", sum, n, x);
}

Flowchart:

or

#include <stdio.h>

void main()
{
	float x,s,t,num=1.00,fac=1.00;
	int i,n,pr,y=2,m=1;

	printf("Input the Value of x :");
	scanf("%f",&x);
	printf("Input the number of terms : ");
	scanf("%d",&n);
	s=1.00; t=1.00;

	for (i=1;i<n;i++)
	{
            for(pr=1;pr<=y;pr++)
                 {
                   fac=fac*pr;
                   num=num*x;

                 }   
          m=m*(-1);
          num=num*m;
          t=num/fac;
	  s=s+t;
          y=y+2;
          num=1.00; 
          fac=1.00;
          
	}
	printf("\nthe sum = %f\nNumber of terms = %d\nvalue of x = %f\n",s,n,x);
} 

Output:

Input the Value of x :2                                                                                       
Input the number of terms : 5                                                                                 
                                                                                                              
the sum = -0.415873                                                                                           
Number of terms = 5                                                                                           
value of x = 2.000000  

Flowchart:

C Programming Code Editor:

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