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

# 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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