`underset("n-digit")ubrace((666…6)) + underset("n-digit")ubrace((888…8))` is equal to :

Options

A. A)`(14(n^2 - 1))`
B. B)`(48)/(9) (10^(2n) -1)`
C. C)`(14)/(9) (10^n -1)`
D. D)none of these

Answer

To solve the problem \( \underset{n-\text{digit}}{\underbrace{666\ldots6}} + \underset{n-\text{digit}}{\underbrace{888\ldots8}} \), we can break it down step by step.

### Step 1: Express the n-digit numbers
The number \( \underset{n-\text{digit}}{\underbrace{666\ldots6}} \) can be expressed as:
\[ 666\ldots6 = 6 \times (10^{n-1} + 10^{n-2} + \ldots + 10^0) \]
This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = 10 \). The number of terms is \( n \).

### Step 2: Sum of the geometric series
The sum of the first \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Applying this formula:
\[ 10^{n-1} + 10^{n-2} + \ldots + 10^0 = \frac{10^n - 1}{10 - 1} = \frac{10^n - 1}{9} \]
Thus,
\[ \underset{n-\text{digit}}{\underbrace{666\ldots6}} = 6 \times \frac{10^n - 1}{9} \]

### Step 3: Express the second n-digit number
Similarly, for \( \underset{n-\text{digit}}{\underbrace{888\ldots8}} \):
\[ 888\ldots8 = 8 \times (10^{n-1} + 10^{n-2} + \ldots + 10^0) = 8 \times \frac{10^n - 1}{9} \]

### Step 4: Combine the two expressions
Now, we combine both expressions:
\[ \underset{n-\text{digit}}{\underbrace{666\ldots6}} + \underset{n-\text{digit}}{\underbrace{888\ldots8}} = 6 \times \frac{10^n - 1}{9} + 8 \times \frac{10^n - 1}{9} \]
Factoring out \( \frac{10^n - 1}{9} \):
\[ = \frac{10^n - 1}{9} (6 + 8) = \frac{10^n - 1}{9} \times 14 \]

### Step 5: Final expression
Thus, the final expression simplifies to:
\[ = \frac{14(10^n - 1)}{9} \]

### Conclusion
The final answer is:
\[ \frac{14(10^n - 1)}{9} \]