<p>The equation \(\displaystyle\sum_{r=1}^{\infty} \dfrac{r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)} = \dfrac{a}{b}\) holds true for co-prime positive integers \(a\) and \(b\). Find \(a+b\).</p>
Step-by-Step Solution
Key Concept: Factorize the denominator as r⁴+r²+1 = (r²+r+1)(r²-r+1) and use partial fraction decomposition to telescope the series into a finite sum.
<p><strong>Step 1: Simplify the numerator</strong></p><p>r³ + (r²+1)² = r³ + r⁴ + 2r² + 1 = r⁴ + r³ + 2r² + 1</p><p><strong>Step 2: Factorize r⁴+r²+1</strong></p><p>r⁴ + r² + 1 = (r²+r+1)(r²-r+1) [key insight: add/subtract r² strategically]</p><p><strong>Step 3: Express numerator in terms of denominator factors</strong></p><p>r⁴ + r³ + 2r² + 1 = (r²+r+1)(r²-r+1) + r³ + r² = (r²+r+1)(r²-r+1) + r(r²+r+1)</p><p>= (r²+r+1)(r²-r+1+r) = (r²+r+1)(r²+1)</p><p><strong>Step 4: Apply partial fractions</strong></p><p>$$\frac{(r^2+r+1)(r^2+1)}{(r^2+r+1)(r^2-r+1) \cdot r(r+1)} = \frac{r^2+1}{r(r+1)(r^2-r+1)}$$</p><p><strong>Step 5: Further decomposition</strong></p><p>$$\frac{r^2+1}{r(r+1)(r^2-r+1)} = \frac{1}{r(r^2-r+1)} - \frac{1}{(r+1)(r^2+r+1)}$$</p><p>This telescopes since r²-r+1 = (r-1)²+(r-1)+1 for the previous term.</p><p><strong>Step 6: Evaluate the telescoping sum</strong></p><p>$$\sum_{r=1}^{\infty} \left[\frac{1}{r(r^2-r+1)} - \frac{1}{(r+1)(r^2+r+1)}\right]$$</p><p>At r=1: 1/(1·1) = 1, and remaining terms vanish as r→∞</p><p>∴ Answer: 1/1, so a=1, b=1, and <strong>a+b = 2</strong></p>
Correct Answer: 1