Sum of the series 1.2.3 + 2.3.+ ... + n(n+1)(n+2) in C

Find the sum up to n terms of the series: 1.2.3 + 2.3.4 + ... + n(n+1)(n+2). In this series, 1.2.3 represents the first term and 2.3.4 represents the second term.

Let's see an example to understand the concept better −

Input: n = 5
Output: 420

Explanation

1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + 5.6.7 = 6 + 24 + 60 + 120 + 210 = 420

The nth term = n(n+1)(n+2); where n = 1, 2, 3, ...

Expanding: n(n+1)(n+2) = n(n² + 3n + 2) = n³ + 3n² + 2n

Using the standard summation formulas −

• ?n = n(n+1)/2
• ?n² = n(n+1)(2n+1)/6
• ?n³ = n²(n+1)²/4

The required sum = ?n³ + 3?n² + 2?n

After algebraic simplification: Sum = n(n+1)(n+2)(n+3)/4

Syntax

sum = n * (n + 1) * (n + 2) * (n + 3) / 4

Method 1: Using Mathematical Formula

This approach directly applies the derived formula for O(1) time complexity −

#include <stdio.h>

int main() {
    int n = 6;
    int sum = n * (n + 1) * (n + 2) * (n + 3) / 4;
    printf("The sum is: %d", sum);
    return 0;
}

The sum is: 756

Method 2: Using Iterative Loop

This approach calculates each term and adds them iteratively −

#include <stdio.h>

int main() {
    int n = 6;
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        sum += i * (i + 1) * (i + 2);
    }
    printf("The sum is: %d", sum);
    return 0;
}

The sum is: 756

Comparison

Conclusion

The series sum n(n+1)(n+2)(n+3)/4 provides an efficient O(1) solution. The iterative method helps understand the series pattern but is slower for large n values.