Sequence Sum Formula Derivation

The given sequence is 8, 88, 888, 8888, ...

We need to find the sum of the first n terms, denoted by S_n.

Step 1: Express the Sequence Terms

The k-th term of the sequence can be written as:

$a_k = \underbrace{88...8}_{k \text{ times}}$

This can be rewritten as:

$a_k = 8 \times \underbrace{11...1}_{k \text{ times}}$

Step 2: Use Powers of 10

The sum of the first k ones, $\underbrace{11...1}_{k \text{ times}}$, can be expressed using the formula $\frac{10^k - 1}{9}$.

Therefore, the k-th term becomes:

$a_k = 8 \times \frac{10^k - 1}{9} = \frac{8}{9}(10^k - 1)$

Step 3: Formulate the Sum S_n

The sum of the first n terms is:

$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} \frac{8}{9}(10^k - 1)$

Factor out the constant $\frac{8}{9}$:

$S_n = \frac{8}{9} \sum_{k=1}^{n} (10^k - 1)$

Separate the sums:

$S_n = \frac{8}{9} \left( \sum_{k=1}^{n} 10^k - \sum_{k=1}^{n} 1 \right)$

Step 4: Apply Geometric Series Formula

The first part, $\sum_{k=1}^{n} 10^k$, is a geometric series with first term $a = 10$, common ratio $r = 10$, and n terms. The sum is $a \frac{r^n - 1}{r - 1} = 10 \frac{10^n - 1}{10 - 1} = \frac{10}{9}(10^n - 1)$.

The second part, $\sum_{k=1}^{n} 1$, is simply n.

Substitute these back into the expression for S_n:

$S_n = \frac{8}{9} \left( \frac{10}{9}(10^n - 1) - n \right)$

Step 5: Simplify the Expression

Distribute the $\frac{8}{9}$:

$S_n = \frac{8}{9} \times \frac{10}{9}(10^n - 1) - \frac{8}{9} \times n$

$S_n = \frac{80}{81}(10^n - 1) - \frac{8}{9}n$

This matches the formula presented in Option 4.