The `n^(th)` term of the series 6+66+666+.....

Options

A. `6/9(10^n-1)`
B. `5xx11^(n-1)`
C. `6xx9^(n-1)`
D. `6xx10^(n-1)`

Answer

To find the \( n^{th} \) term of the series \( 6 + 66 + 666 + \ldots \), we can observe the pattern in the terms of the series.

### Step-by-Step Solution:

1. **Identify the Pattern**:
- The first term is \( 6 \).
- The second term is \( 66 \) (which can be written as \( 6 \times 11 \)).
- The third term is \( 666 \) (which can be written as \( 6 \times 111 \)).
- We can see that each term consists of the digit \( 6 \) repeated \( n \) times.

2. **Express the \( n^{th} \) Term**:
- The \( n^{th} \) term can be expressed as \( 6 \) followed by \( n \) digits of \( 6 \).
- This can be mathematically represented as \( 6 \times (10^{n-1} + 10^{n-2} + \ldots + 10^0) \).

3. **Sum of the Geometric Series**:
- The expression \( 10^{n-1} + 10^{n-2} + \ldots + 10^0 \) is a geometric series.
- The sum of a geometric series can be calculated using the formula:
\[ S_n = a \frac{(r^n - 1)}{(r - 1)} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.
- Here, \( a = 1 \), \( r = 10 \), and there are \( n \) terms.
- Thus, the sum becomes:
\[ S_n = 1 \cdot \frac{(10^n - 1)}{(10 - 1)} = \frac{10^n - 1}{9} \]

4. **Substituting Back**:
- Now substituting back into our expression for the \( n^{th} \) term:
\[ T_n = 6 \times \frac{10^n - 1}{9} \]
- This simplifies to:
\[ T_n = \frac{6}{9} \times (10^n - 1) = \frac{2}{3} \times (10^n - 1) \]
- However, we can also express it as:
\[ T_n = \frac{6}{9} \times 10^n - \frac{6}{9} = \frac{6 \times 10^n - 6}{9} \]

5. **Final Form**:
- The final form of the \( n^{th} \) term is:
\[ T_n = \frac{6}{9} \times (10^n - 1) = \frac{2}{3} \times (10^n - 1) \]

### Conclusion:
Thus, the \( n^{th} \) term of the series \( 6 + 66 + 666 + \ldots \) is given by:
\[ T_n = \frac{6}{9} \times (10^n - 1) = \frac{2}{3} \times (10^n - 1) \]