Find the sum of n terms of the sequence 8, 88, 888, 8888,...

Steps to Solve

1. Identify the Sequence Pattern
   The sequence given is: 8, 88, 888, 8888, etc. Each term can be represented as $T_n = 8 \times (\underbrace{10^0 + 10^1 + \ldots + 10^{n-1}}_{n \text{ digits of } 8})$.

2. Sum Formula for Each Term
   Using the formula for the sum of a geometric series, we can express the $n$-th term as: $$ T_n = 8 \times \frac{10^n - 1}{9} $$
   For $n$ terms, the last term will be $T_n$.

3. Sum of the First n Terms
   To find the sum ( S_n ) of the first $n$ terms: $$ S_n = T_1 + T_2 + T_3 + \ldots + T_n $$ Substituting for $T_n$ using the earlier expression: $$ S_n = \sum_{k=1}^{n} 8 \times \frac{10^k - 1}{9} $$

4. Simplifying the Sum
   This can be simplified as: $$ S_n = \frac{8}{9} \left( \sum_{k=1}^{n} 10^k - n \right) $$ The sum of the first $n$ terms of the powers of 10 is: $$ \sum_{k=1}^{n} 10^k = 10 \frac{10^n - 1}{9} $$

5. Final Expression for the Sum
   Combining all parts, we get: $$ S_n = \frac{8}{9} \left( \frac{10(10^n - 1)}{9} - n \right) $$ Now, this can be calculated for any specific value of $n$.