Method 1: The Intuitive Approach (Substitution)
The most intuitive way to solve this is to test the general formula against a concrete, easily calculable reality. If an option truly represents the sum of $n$ terms, it must be correct for $n = 1$ and $n = 2$.
Test Case 1: $n = 1$
If we only want the sum of the first term ($n = 1$), our sum is simply:
$$S_1 = 4$$
Let's plug $n = 1$ into the given options to see which one yields $4$:
A. $\frac{4}{81} [10^2 - 9(1) - 1] = \frac{4}{81} [100 - 10] = \frac{4}{81} [90] \neq 4$
B. $\frac{4}{81} [10^0 - 9(1) - 1] = \frac{4}{81} [1 - 10] = \frac{4}{81} [-9] \neq 4$
C. $\frac{4}{81} [10^2 - 9(1) - 10] = \frac{4}{81} [100 - 19] = \frac{4}{81} [81] = 4 \times 1 = \mathbf{4}$ (This matches!)
D. $\frac{4}{81} [10^1 - 9(1) - 10] = \frac{4}{81} [10 - 19] = \frac{4}{81} [-9] \neq 4$
Test Case 2: $n = 2$ (Optional Verification)
To be absolutely certain, let's test the sum of the first two terms:
$$S_2 = 4 + 44 = 48$$
Plugging $n = 2$ into our winning Option C:
$\frac{4}{81} [10^3 - 9(2) - 10]$
$\frac{4}{81} [1000 - 18 - 10]$
$\frac{4}{81} [972]$
$4 \times 12 = \mathbf{48}$
The formula holds perfectly. The answer is Option C.
Method 2: First Principles (Algebraic Derivation)
To derive this algebraically, we need to manipulate the series into a Geometric Progression (GP) that we know how to sum.
Step 1: Set up the sequence and factor out the 4.
$$S_n = 4 + 44 + 444 + \dots \text{ to } n \text{ terms}$$
$$S_n = 4(1 + 11 + 111 + \dots \text{ to } n \text{ terms})$$
Step 2: Multiply and divide by 9.
We do this to create values that are close to powers of 10, which allows us to form a GP.
$$S_n = \frac{4}{9}(9 + 99 + 999 + \dots \text{ to } n \text{ terms})$$
Step 3: Rewrite terms as powers of 10 minus 1.
$$S_n = \frac{4}{9} [(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$$
Step 4: Regroup the terms.
Separate the powers of 10 from the $1$s.
$$S_n = \frac{4}{9} [(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots \text{ to } n \text{ terms})]$$
Step 5: Apply the GP sum formula.
The first bracket is a Geometric Progression where the first term $a = 10$ and the common ratio $r = 10$. The sum of a GP is $S = a \frac{(r^n - 1)}{(r - 1)}$.
The second bracket is simply $n$ (since $1$ is added $n$ times).
$$S_n = \frac{4}{9} \left[ \frac{10(10^n - 1)}{10 - 1} - n \right]$$
$$S_n = \frac{4}{9} \left[ \frac{10}{9}(10^n - 1) - n \right]$$
Step 6: Simplify to match the options.
Find a common denominator of 9 inside the brackets to factor it out.
$$S_n = \frac{4}{9} \left[ \frac{10(10^n - 1) - 9n}{9} \right]$$
$$S_n = \frac{4}{81} [10(10^n) - 10 - 9n]$$
$$S_n = \frac{4}{81} [10^{n+1} - 9n - 10]$$
This derivation directly matches Option C.
