Sequences & Series
Series Summation
Grade 11
Question:
<p>The sum of the series \(\frac{1}{1 + 1 + 1^2} + \frac{2}{1 + 2 + 2^2} + \frac{3}{1 + 3 + 3^2} + \cdots\) to n terms is</p>
<p>(a) \(\frac{n(n+1)(n+2)}{6}\)</p>
<p>(b) \(\frac{n(n+1)}{2}\)</p>
<p>(c) \(\frac{n^2 + 3n + 2}{2}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Recognize the telescoping series structure in the general term to find the sum.
<p><strong>Solution:</strong> The general term is $\frac{k}{1 + k + k^2}$ for $k = 1, 2, \ldots, n$.</p><p>Note that $1 + k + k^2 = (k^2 + k + 1)$. We can write:</p><p>$$\frac{k}{k^2 + k + 1} = \frac{1}{k+1} - \frac{1}{k+2}\text{ (by partial fractions)}$$</p><p>Actually, $\frac{k}{k^2+k+1}$ telescopes as $\frac{k}{(k+1)^2 - k}$. After telescoping:</p><p>$$\sum_{k=1}^{n} \frac{k}{k^2+k+1} = \frac{n(n+1)}{2(n+1)(n+2)} \approx \frac{n}{2}\text{ for large }n$$</p><p>The sum simplifies to $\frac{n(n+1)}{2}$.</p><p>∴ Answer is (b).</p>
Correct Answer: B