Let $S$ be the sum of the first nine terms of the sequence \[x+a, x^2+2a, x^3+3a, \cdots.\]
Then $S$ equals:
$ \textbf{(A)}\ \frac{50a+x+x^8}{x+1} \qquad\textbf{(B)}\ 50a-\frac{x+x^{10}}{x-1}\qquad\textbf{(C)}\ \frac{x^9-1}{x+1}+45a\qquad$$\textbf{(D)}\ \frac{x^{10}-x}{x-1}+45a\qquad\textbf{(E)}\ \frac{x^{11}-x}{x-1}+45a$
1. The given sequence is \( x + a, x^2 + 2a, x^3 + 3a, \ldots \). We need to find the sum of the first nine terms of this sequence.
2. Let's denote the \( n \)-th term of the sequence by \( T_n \). The general form of the \( n \)-th term is \( T_n = x^n + na \).
3. The sum of the first nine terms \( S \) can be written as:
   \[
   S = (x + a) + (x^2 + 2a) + (x^3 + 3a) + \cdots + (x^9 + 9a)
   \]
4. We can separate the sum into two parts: the sum of the geometric series and the sum of the arithmetic series:
   \[
   S = (x + x^2 + x^3 + \cdots + x^9) + (a + 2a + 3a + \cdots + 9a)
   \]
5. The first part is the sum of a geometric series with the first term \( x \) and common ratio \( x \):
   \[
   \sum_{k=1}^{9} x^k = x + x^2 + x^3 + \cdots + x^9
   \]
   The sum of this geometric series is given by:
   \[
   \sum_{k=1}^{9} x^k = x \frac{x^9 - 1}{x - 1}
   \]
6. The second part is the sum of an arithmetic series with the first term \( a \) and common difference \( a \):
   \[
   \sum_{k=1}^{9} ka = a(1 + 2 + 3 + \cdots + 9)
   \]
   The sum of the first \( n \) natural numbers is given by:
   \[
   \sum_{k=1}^{n} k = \frac{n(n+1)}{2}
   \]
   Therefore:
   \[
   \sum_{k=1}^{9} k = \frac{9 \cdot 10}{2} = 45
   \]
   Hence:
   \[
   \sum_{k=1}^{9} ka = a \cdot 45 = 45a
   \]
7. Combining both parts, we get:
   \[
   S = x \frac{x^9 - 1}{x - 1} + 45a
   \]
8. Simplifying the expression, we get:
   \[
   S = \frac{x^{10} - x}{x - 1} + 45a
   \]

The final answer is \(\boxed{\frac{x^{10} - x}{x - 1} + 45a}\)