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IP-03.c

Problem Statement

Write a program to compute the sum of the first n terms of the series using a function: S=1−2+3−4+5−6+…

Metadata

Property Detail
Author Amit Dutta (amitdutta4255@gmail.com)
License MIT
Difficulty Beginner (index: 3 / 10)

Concepts

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  • Recursion
  • Iteration

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Source Code

#include <stdio.h>

int sum_of_series(int);

int main()
{
    int n;
    printf("Enter the n: ");
    scanf("%d", &n);
    printf("\nSum of the first %d terms of the series = %d", n, sum_of_series(n));
    return 0;
}

int sum_of_series(int n)
{
    int sum = 0;
    int i;
    for (i = 1; i <= n; i++)
    {
        if (i % 2 == 0)
        {
            sum -= i;
        }
        else
        {
            sum += i;
        }
    }
    return sum;
}

Explanation

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Copy the prompt below and paste it into any AI assistant.

    You are explaining a C programming code to a beginner.

    STRICT RULES:

    - Only use the given code. Do NOT assume anything not present.

    - Do NOT add extra examples.

    - Keep explanation clear and short.

    - If something is unclear, say "Not clear from code".

    - Follow the exact format below. Do NOT change headings.

    FORMAT:

    [START]

    ## What it does

    (Explain the overall purpose in 1-2 sentences)

    ## Step-by-step

    (Explain how the code works in steps, simple language)

    ## Key Concepts

    (List concepts like loop, condition, function, etc.)

    ## Notes

    (Mention any limitations, errors, or assumptions)

    [END]

    CODE (IP-03.c):

    #include <stdio.h>

    int sum_of_series(int);

    int main()
    {
        int n;
        printf("Enter the n: ");
        scanf("%d", &n);
        printf("\nSum of the first %d terms of the series = %d", n, sum_of_series(n));
        return 0;
    }

    int sum_of_series(int n)
    {
        int sum = 0;
        int i;
        for (i = 1; i <= n; i++)
        {
            if (i % 2 == 0)
            {
                sum -= i;
            }
            else
            {
                sum += i;
            }
        }
        return sum;
    }