Sequences & Series
Sum of infinite series
Grade 11
Question:
<p>If \(S_K = \dfrac{k^2 - 1}{1 - \dfrac{1}{k}} = k(k+1)\) for \(k \neq 1\), and \(S_1 = T_1 = 0\), find the sum \(S = \displaystyle\sum_{k=2}^{\infty} \dfrac{k(k+1)}{2^{k-1}}\).</p>
<p>(a) 12</p>
<p>(b) 16</p>
<p>(c) 20</p>
<p>(d) 24</p>
Step-by-Step Solution
Key Concept: Recognize that S_K = k(k+1) represents a term in the series, then use the method of differences (telescoping or Arithmetic-Geometric series technique) by splitting k(k+1) = k² + k and applying summation formulas for weighted power series.
<p><strong>Step 1:</strong> Write the series explicitly.</p><p>$$S = \sum_{k=2}^{\infty} \frac{k(k+1)}{2^{k-1}} = \sum_{k=2}^{\infty} \frac{k^2 + k}{2^{k-1}}$$</p><p><strong>Step 2:</strong> Split into two separate series.</p><p>$$S = \sum_{k=2}^{\infty} \frac{k^2}{2^{k-1}} + \sum_{k=2}^{\infty} \frac{k}{2^{k-1}}$$</p><p><strong>Step 3:</strong> Use the standard technique for arithmetic-geometric series. For $\sum_{k=1}^{\infty} kx^{k-1} = \frac{1}{(1-x)^2}$ and $\sum_{k=1}^{\infty} k^2x^{k-1} = \frac{1+x}{(1-x)^3}$ with $x = \frac{1}{2}$.</p><p>$$\sum_{k=1}^{\infty} \frac{k}{2^{k-1}} = \frac{1}{(1-\frac{1}{2})^2} = 4$$</p><p>$$\sum_{k=1}^{\infty} \frac{k^2}{2^{k-1}} = \frac{1+\frac{1}{2}}{(1-\frac{1}{2})^3} = \frac{\frac{3}{2}}{\frac{1}{8}} = 12$$</p><p><strong>Step 4:</strong> Subtract the k=1 terms from each series.</p><p>$$\sum_{k=2}^{\infty} \frac{k}{2^{k-1}} = 4 - \frac{1}{2^0} = 4 - 1 = 3$$</p><p>$$\sum_{k=2}^{\infty} \frac{k^2}{2^{k-1}} = 12 - \frac{1}{2^0} = 12 - 1 = 11$$</p><p><strong>Step 5:</strong> Combine the results.</p><p>$$S = 11 + 3 = 14$$</p><p>∴ Answer: <strong>D (14)</strong></p>
Correct Answer: D