Sequences & Series
Geometric Progression
Grade 11
Question:
<p>If \(a_1, a_2, a_3, \ldots\) are in GP with first term \(a\) and common ratio \(r\), then \(\frac{a_1 a_2}{a_1^2 + a_2^2} + \frac{a_2 a_3}{a_2^2 + a_3^2} + \frac{a_3 a_4}{a_3^2 + a_4^2} + \cdots + \frac{a_n a_{n+1}}{a_n^2 + a_{n+1}^2}\) is equal to</p>
<p>(a) \(\frac{nr}{1 + r^2}\)</p>
<p>(b) \(\frac{(n + 1)r}{1 + r^2}\)</p>
<p>(c) \(\frac{nr}{1 + r}\)</p>
<p>(d) \(\frac{(n + 1)r}{1 + r}\)</p>
Step-by-Step Solution
Key Concept: Express each term using the GP property (a_{n+1} = a_n · r) and find a telescoping pattern or simplification. Use partial fractions to decompose each denominator term.
<p><strong>Step 1: Express terms using GP property</strong></p><p>Since a_k = ar^{k-1}, we have a_{k+1} = ar^k = a_k · r.</p><p>For the k-th term: a_k a_{k+1} = a_k · (a_k r) = a_k^2 r</p><p>Also: a_{k+1}^2 = (a_k r)^2 = a_k^2 r^2</p><p></p><p><strong>Step 2: Simplify the general term</strong></p><p>$$\frac{a_k a_{k+1}}{a_k^2 + a_{k+1}^2} = \frac{a_k^2 r}{a_k^2 + a_k^2 r^2} = \frac{a_k^2 r}{a_k^2(1 + r^2)} = \frac{r}{1 + r^2}$$</p><p></p><p><strong>Step 3: Notice independence from k</strong></p><p>Remarkably, each term equals the same value $\frac{r}{1 + r^2}$, independent of which term in the sequence it is.</p><p></p><p><strong>Step 4: Sum all n terms</strong></p><p>Since there are n terms in the sum (from k=1 to k=n), we have:</p><p>$$\text{Sum} = n \cdot \frac{r}{1 + r^2} = \frac{nr}{1 + r^2}$$</p><p></p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A