C Programming for Sum of sequence 2, 22, 222, .........

Given is a sequence: 2, 22, 222, 2222… and we need to find the sum of this sequence. We need to derive a mathematical formula to calculate the sum efficiently.

Syntax

sum = 2 * (pow(10, n) - 1 - (9 * n)) / 81;

Mathematical Derivation

The formula derivation follows these steps −

sum = [2 + 22 + 222 + 2222 + ...]
sum = 2 * [1 + 11 + 111 + 1111 + ...]
sum = 2/9 * [9 + 99 + 999 + 9999 + ...]
sum = 2/9 * [(10-1) + (100-1) + (1000-1) + (10000-1) + ...]
sum = 2/9 * [10 + 10² + 10³ + 10? + ... - n]
sum = 2/9 * [(10^(n+1) - 10)/9 - n]
sum = 2 * (10^(n+1) - 10 - 9n) / 81

Example

Here's a complete program to calculate the sum of the sequence for the first n terms −

#include <stdio.h>
#include <math.h>

int main() {
    int n = 3;
    float sum = 2 * (pow(10, n + 1) - 10 - (9 * n)) / 81;
    printf("Sum of first %d terms: %.0f<br>", n, sum);
    
    // Verify by calculating manually
    printf("Manual verification: 2 + 22 + 222 = %d<br>", 2 + 22 + 222);
    
    return 0;
}

Sum of first 3 terms: 246
Manual verification: 2 + 22 + 222 = 246

Example with User Input

Here's an interactive version that accepts user input −

#include <stdio.h>
#include <math.h>

int main() {
    int n = 5;
    float sum = 2 * (pow(10, n + 1) - 10 - (9 * n)) / 81;
    
    printf("Sequence: ");
    for(int i = 1; i <= n; i++) {
        int term = 2 * (pow(10, i) - 1) / 9;
        printf("%d", term);
        if(i < n) printf(" + ");
    }
    printf("<br>");
    
    printf("Sum of first %d terms: %.0f<br>", n, sum);
    return 0;
}

Sequence: 2 + 22 + 222 + 2222 + 22222
Sum of first 5 terms: 24690

Key Points

• The formula uses geometric series properties to derive an efficient O(1) solution.
• Each term in the sequence follows the pattern: 2 * (10^i - 1) / 9 for the i-th term.
• The pow() function from math.h is used for exponentiation.

Conclusion

The mathematical formula 2 * (10^(n+1) - 10 - 9n) / 81 provides an efficient way to calculate the sum of the sequence 2, 22, 222, ... without iterating through each term individually.