How to Write a C Program to Compute the Series Using a While Loop

When you need to compute a series in a C program, a while loop can be a powerful tool. In this tutorial, we will guide you through the process of computing the following series:

23 × 53 × 83 × n3(), which is generated by the sequence 2, 5, 8, (3k 2).

What You Will Learn:

Understanding the mathematical series. Constructing a C program using a while loop to compute the series. Using closed-form expressions for large values. Implementing a Python example for both closed-form and loop-based solutions.

Understanding the Mathematical Series

The series we need to compute can be described as:

S 23 × 53 × 83 × n3

where each term in the sequence is given by the formula: 3k 2, starting with k 0.

To solve this problem, we can use a while loop in C to iterate through the series until we reach or exceed the value of n.

Constructing the C Program

Here is a sample C program that performs the computation:

#include ltstdio.hgt#include ltmath.hgtint main() {    int n;    printf("Enter the value of n: ");    scanf("%d", n);    int sum  0;    int term  2;  // Start with the first term in the series 2    // Loop until the term exceeds n    while (term lt n) {        sum  pow(term, 3);  // Add the cube of the term to the sum        term  3 * term   2;  // Move to the next term in the series    }    printf("The computed sum is: %d
", sum);    return 0;}

Explanation

Input: The program prompts the user to enter the value of n. Initialization: - n - The value of n is stored. - sum - Initialized to 0 to store the total sum of the series. - term - Initialized to 2, which is the first term of the series. While Loop: - The loop continues as long as term lt n. - Inside the loop, the cube of term is added to sum. - term is then incremented by 3 to get the next term in the series. Output: Finally, the program prints the computed sum.

Example Usage

If the user inputs n 10, the output will be the sum of 2^3 and 5^3 as 8^3 exceeds 10, resulting in 8^3 5^3 125 8 133.

Series Summation and Closed-Form Expressions

For a low value n 3kmax - 1, a while or for loop is efficient. Larger values can benefit from summation notation and known closed form sums.

The series can be represented as:

[ S sum_{k1}^{n} (3k - 1)^3 2^3 times 5^3 times 8^3 times ldots times n^3 ]

This can be expanded as:

[ S sum_{k1}^{n} (27k^3 - 27k^2 9k - 1) ]

Distributing the summation sign to make three summations, we get:

[ S 27 sum_{k1}^{n} k^3 - 27 sum_{k1}^{n} k^2 9 sum_{k1}^{n} k - sum_{k1}^{n} 1 ]

The rightmost sum is clearly n. The other summations have known closed-form sums to replace them.

Sum of n, n2, or n3 - Brilliant Math and Science Wiki

[ S frac{9}{4}n(n-1)^2 - 27 cdot frac{n(n-1)(2n-1)}{6} 9 cdot frac{n(n-1)}{2} - n ]

Simplify this to get the closed form sum for S.

[ S frac{9}{4}n(n-1)(3n^2 - n - 2) - n ]

[ S frac{9}{4}n(n-1)(3n^2 - 3n - 4) - n ]

[ S frac{9}{4}n(n-1)(3n^2 - n) - n ]

Python Implementation

This Python example shows both the closed-form and loop-based solutions:

def summation_closed_form(n):    return (9 * n * (n - 1) / 4 - 27 * n * (n - 1) * (2 * n - 1) / 6   9 * n * (n - 1) / 2 - n)def sum_loop(n):    add_up  0    for k in range(1, n   1):        add_up   (3 * k - 1) ** 3    return add_upif __name__  '__main__':    num  10  # Example n value    print(f"Closed form result: {summation_closed_form(num)}")    print(f"Loop result: {sum_loop(num)}")    assert summation_closed_form(num)  sum_loop(num)

We can see that both methods give the same result, confirming the correctness of the closed-form expression.