<p>If \(\dfrac{x^2 + x + 1}{1 - x} = a_0 + a_1 x + a_2 x^2 + \cdots\), then \(\displaystyle\sum_{r=1}^{50} a_r\) is equal to</p>
Step-by-Step Solution
Key Concept: Expand the LHS using the geometric series formula after polynomial division, then extract coefficients by comparing powers of x. The sum of coefficients (except a₀) can be found by evaluating at a strategic value of x.
<p><strong>Step 1:</strong> Rewrite the equation by expanding the LHS using geometric series.</p><p>$$\frac{x^2 + x + 1}{1-x} = (x^2 + x + 1) \cdot \frac{1}{1-x} = (x^2 + x + 1)(1 + x + x^2 + x^3 + \cdots)$$</p><p><strong>Step 2:</strong> Multiply out the series:</p><p>$$= 1 + x + x^2 + x + x^2 + x^3 + x^2 + x^3 + x^4 + \cdots$$</p><p>$$= 1 + 2x + 3x^2 + 3x^3 + 3x^4 + \cdots$$</p><p>So: $a_0 = 1$, $a_1 = 2$, and $a_r = 3$ for all $r \geq 2$</p><p><strong>Step 3:</strong> Calculate the sum:</p><p>$$\sum_{r=1}^{50} a_r = a_1 + \sum_{r=2}^{50} a_r = 2 + 3(49) = 2 + 147 = 149$$</p><p>∴ Answer: <strong>C</strong> (149)</p>
Correct Answer: C