If `(x^(2)+x+1)/(1-x) = a_(0) + a_(1)x+a_(2)x^(2)+"…."`, then `sum_(r=1)^(50) a_(r)` equal to

Options

A. 148
B. 146
C. 149
D. none of these

Explanation

To solve the problem, we need to analyze the given expression and find the coefficients \( a_r \) for the series expansion. Let's break it down step by step.

### Step 1: Rewrite the Expression
We start with the expression:

\[ \frac{x^2 + x + 1}{1 - x} \]

We can separate the numerator:

\[ \frac{x^2}{1 - x} + \frac{x}{1 - x} + \frac{1}{1 - x} \]

### Step 2: Expand Each Term
Now, we will expand each term using the formula for the geometric series:

\[ \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n \]

Thus, we can write:

1. For \( \frac{1}{1 - x} \):
\[ \frac{1}{1 - x} = 1 + x + x^2 + x^3 + \ldots \]

2. For \( \frac{x}{1 - x} \):
\[ \frac{x}{1 - x} = x + x^2 + x^3 + \ldots \]

3. For \( \frac{x^2}{1 - x} \):
\[ \frac{x^2}{1 - x} = x^2 + x^3 + x^4 + \ldots \]

### Step 3: Combine the Series
Now, we combine all these series:

\[ \frac{x^2 + x + 1}{1 - x} = (1 + x + x^2 + x^3 + \ldots) + (x + x^2 + x^3 + \ldots) + (x^2 + x^3 + \ldots) \]

Combining like terms, we get:

- The constant term \( a_0 = 1 \)
- The coefficient of \( x \) (from \( 1 \) and \( x \)) gives \( a_1 = 2 \)
- The coefficient of \( x^2 \) gives \( a_2 = 3 \)
- The coefficient of \( x^3 \) gives \( a_3 = 3 \)
- The coefficient of \( x^4 \) gives \( a_4 = 3 \)
- For \( r \geq 3 \), \( a_r = 3 \)

### Step 4: Find the Sum \( \sum_{r=1}^{50} a_r \)
Now we calculate the sum:

\[ \sum_{r=1}^{50} a_r = a_1 + a_2 + a_3 + a_4 + \ldots + a_{50} \]

Substituting the values we found:

\[ = 2 + 3 + 3 + 3 + \ldots + 3 \]

Here, \( a_3 \) to \( a_{50} \) are all equal to \( 3 \). There are \( 48 \) terms from \( a_3 \) to \( a_{50} \).

Calculating the sum:

\[ = 2 + 3 + 3 \times 48 \]
\[ = 2 + 3 + 144 \]
\[ = 2 + 147 = 149 \]

### Final Answer
Thus, the value of \( \sum_{r=1}^{50} a_r \) is:

\[ \boxed{149} \]