Let `S_n` denotes the sum of the terms of n series `(1lt=nlt=9)` `1+22+333+.....999999999`, is

Options

A. `S_(n)-S_(n-1)=(1)/(9)(10^(n)-n^(2)+n)`
B. `S_(n)=(1)/(9)(10^(n)-n^(2)+2n-2)`
C. `9(S_(n)-S_(n-1))=n(10^(n)-1)`
D. None of these

Explanation

To solve the problem, we need to find the sum \( S_n \) of the series \( 1 + 22 + 333 + \ldots + \underbrace{nnn\ldots n}_{n \text{ times}} \) for \( n \) ranging from 1 to 9.

### Step-by-Step Solution:

1. **Understanding the Series**:
The series can be expressed as:
\[ S_n = 1 + 22 + 333 + \ldots + \underbrace{nnn\ldots n}_{n \text{ times}} \]
Each term \( k \) (where \( k \) ranges from 1 to \( n \)) contributes \( k \) repeated \( k \) times.

2. **Expressing Each Term**:
The \( k \)-th term can be expressed as:
\[ k \times (10^{k-1} + 10^{k-2} + \ldots + 10^0) = k \times \frac{10^k - 1}{9} \]
This is because the sum of a geometric series \( 10^{k-1} + 10^{k-2} + \ldots + 10^0 \) equals \( \frac{10^k - 1}{10 - 1} \).

3. **Calculating \( S_n \)**:
Therefore, we can write:
\[ S_n = \sum_{k=1}^{n} k \times \frac{10^k - 1}{9} \]
This can be simplified to:
\[ S_n = \frac{1}{9} \sum_{k=1}^{n} k (10^k - 1) \]

4. **Splitting the Summation**:
We can split the summation:
\[ S_n = \frac{1}{9} \left( \sum_{k=1}^{n} k \cdot 10^k - \sum_{k=1}^{n} k \right) \]
The second summation \( \sum_{k=1}^{n} k \) is the sum of the first \( n \) natural numbers, which equals \( \frac{n(n+1)}{2} \).

5. **Finding \( \sum_{k=1}^{n} k \cdot 10^k \)**:
To find \( \sum_{k=1}^{n} k \cdot 10^k \), we can use the formula:
\[ \sum_{k=1}^{n} k x^k = x \frac{d}{dx} \left( \sum_{k=0}^{n} x^k \right) = x \frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) \]
Evaluating this for \( x = 10 \) gives us:
\[ \sum_{k=1}^{n} k \cdot 10^k = 10 \frac{d}{d10} \left( \frac{1 - 10^{n+1}}{1 - 10} \right) \]

6. **Final Expression for \( S_n \)**:
After calculating the above derivatives and substituting back, we can find \( S_n \) in terms of \( n \).

7. **Evaluating for \( n = 9 \)**:
Finally, we can substitute \( n = 9 \) into our expression for \( S_n \) to find the required sum.