C++ Program To Find Sum Of Series 1 1 2 1 3

In this article, you will learn how to write a C++ program to find the sum of a specific non-standard series: 1, 1, 2, 1, 3, ... up to a given number of terms. We will explore the pattern within this series and implement an efficient solution.

Problem Statement

The challenge is to calculate the sum of the first 'n' terms of the series that begins with 1, 1, 2, 1, 3. Upon closer inspection, the pattern reveals that terms at odd positions follow an incrementing sequence (1, 2, 3, ...) while terms at even positions are consistently 1.

Specifically:

• For an odd position i, the term is (i + 1) / 2.
• For an even position i, the term is 1.

The objective is to prompt the user for the value of 'n' and then compute the total sum of the series up to that 'n'.

Example

Let's illustrate the series and its sum for a few terms:

• For n = 1: The series is 1. Sum = 1.
• For n = 2: The series is 1, 1. Sum = 2.
• For n = 3: The series is 1, 1, 2. Sum = 4.
• For n = 5: The series is 1, 1, 2, 1, 3. Sum = 8.

Background & Knowledge Prerequisites

To understand and implement the solution in this article, readers should have a basic understanding of:

• C++ syntax: Variables, data types, basic arithmetic operators.
• Control flow: for loops and if-else conditional statements.
• Input/Output operations: Using std::cin for input and std::cout for output.
• Modulo operator (%): Used to determine if a number is odd or even.

Use Cases or Case Studies

This type of problem, involving pattern recognition and conditional summation, is common in various scenarios:

• Mathematical challenges: Identifying and computing sums of sequences with specific rules.
• Programming contests: A frequent type of problem designed to test logical thinking and loop implementation.
• Educational exercises: Helps reinforce understanding of loops, conditionals, and basic arithmetic in programming.
• Data sequence generation: Understanding how to generate and process sequences based on positional rules.

Solution Approaches

Iterative Summation with Conditional Logic

This approach involves iterating from 1 up to the specified number of terms n. Within each iteration, we determine if the current term's position is odd or even using the modulo operator. Based on this, we calculate the term's value and add it to a running total.

One-line summary: Iterate through each position, apply the series rule based on odd/even position, and accumulate the sum.

// Sum of Series 1 1 2 1 3 ...
#include <iostream>
using namespace std;

int main() {
    // Step 1: Declare variables for the number of terms and the sum
    int n;
    long long sum = 0; // Use long long for sum to prevent potential overflow for large 'n'

    // Step 2: Prompt user for the number of terms
    cout << "Enter the number of terms (n) for the series: ";
    cin >> n;

    // Step 3: Input validation (optional, but good practice)
    if (n <= 0) {
        cout << "Please enter a positive integer for n." << endl;
        return 1; // Indicate an error
    }

    // Step 4: Loop through each term from 1 to n
    for (int i = 1; i <= n; ++i) {
        // Step 5: Check if the current position 'i' is odd or even
        if (i % 2 != 0) { // If 'i' is odd
            sum += (i + 1) / 2;
        } else { // If 'i' is even
            sum += 1;
        }
    }

    // Step 6: Display the calculated sum
    cout << "The sum of the series up to " << n << " terms is: " << sum << endl;

    return 0;
}

Run

Sample Output:

Enter the number of terms (n) for the series: 5
The sum of the series up to 5 terms is: 8

Enter the number of terms (n) for the series: 10
The sum of the series up to 10 terms is: 17

Stepwise Explanation for Clarity:

1. Initialization: We declare an integer n to store the number of terms and a long long variable sum initialized to 0. long long is chosen for sum to accommodate potentially large sums, as the series grows with n.
2. User Input: The program prompts the user to enter the value for n and stores it.
3. Input Validation: A basic check ensures n is a positive integer. This prevents errors or unexpected behavior for invalid inputs.
4. Looping: A for loop iterates from i = 1 to n. Each i represents the position of a term in the series.
5. Conditional Term Calculation:
   • Inside the loop, i % 2 != 0 checks if i is an odd number.

1. Display Result: After the loop completes, the final calculated sum is printed to the console.

Conclusion

Finding the sum of a non-standard series like 1, 1, 2, 1, 3, ... requires careful observation to identify the underlying pattern. By breaking down the series based on the parity (odd or even) of the term's position, we can apply conditional logic within a simple iterative loop to accurately calculate the sum. This approach is straightforward and efficient for computing sums of such sequences.

Summary

• The series 1, 1, 2, 1, 3, ... follows a pattern where odd-positioned terms are (position + 1) / 2 and even-positioned terms are 1.
• An iterative for loop is used to process each term up to n.
• The modulo operator (%) helps determine if a term's position is odd or even.
• Conditional if-else statements apply the correct rule for calculating each term's value.
• A running sum variable accumulates these values to provide the final result.