<p>If the sum to infinity of the series \(3 + (3 + d)\dfrac{1}{4} + (3 + 2d)\dfrac{1}{4^2} + \cdots \infty\) is \(\dfrac{44}{9}\), then find \(d\).</p>
Step-by-Step Solution
Key Concept: Split the series into two parts: a geometric series (constant terms) and a geometric series multiplied by an arithmetic sequence (variable terms), then sum each separately using the formula for infinite geometric series and its derivative.
<p><strong>Step 1:</strong> Separate the series into two parts:</p><p>$$S = 3 + (3+d)\frac{1}{4} + (3+2d)\frac{1}{4^2} + (3+3d)\frac{1}{4^3} + \cdots$$</p><p>$$S = \underbrace{\left(3 + 3\cdot\frac{1}{4} + 3\cdot\frac{1}{4^2} + \cdots\right)}_{S_1} + \underbrace{\left(d\cdot\frac{1}{4} + 2d\cdot\frac{1}{4^2} + 3d\cdot\frac{1}{4^3} + \cdots\right)}_{S_2}$$</p><p><strong>Step 2:</strong> Calculate $S_1$ (geometric series with first term 3 and ratio $\frac{1}{4}$):</p><p>$$S_1 = \frac{3}{1-\frac{1}{4}} = \frac{3}{\frac{3}{4}} = 4$$</p><p><strong>Step 3:</strong> Calculate $S_2$ using the formula for $\sum_{n=1}^{\infty} nrx^{n-1} = \frac{1}{(1-x)^2}$:</p><p>$$S_2 = d\left(\frac{1}{4} + 2\cdot\frac{1}{4^2} + 3\cdot\frac{1}{4^3} + \cdots\right) = d \cdot \frac{\frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}$$</p><p>$$S_2 = d \cdot \frac{\frac{1}{4}}{\frac{9}{16}} = d \cdot \frac{1}{4} \cdot \frac{16}{9} = \frac{4d}{9}$$</p><p><strong>Step 4:</strong> Set up the equation:</p><p>$$S = S_1 + S_2 = 4 + \frac{4d}{9} = \frac{44}{9}$$</p><p>$$\frac{4d}{9} = \frac{44}{9} - 4 = \frac{44-36}{9} = \frac{8}{9}$$</p><p>$$4d = 8$$</p><p><strong>∴ Answer: $d = 2$</strong></p>
Correct Answer: 2