The general term is \(\frac{\sum_{k=1}^{n}k^3}{\sum_{k=1}^{n}(2k-1)}\). We know \(\sum_{k=1}^{n}k^3 = \left(\frac{n(n+1)}{2}\right)^2\) and \(\sum_{k=1}^{n}(2k-1) = n^2\). So the nth term is \(\frac{[n(n+1)/2]^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4}\). Sum of first 9 terms = \(\sum_{n=1}^{9}\frac{(n+1)^2}{4} = \frac{1}{4}\sum_{n=1}^{9}(n+1)^2 = \frac{1}{4}\sum_{m=2}^{10}m^2 = \frac{1}{4}\left[\sum_{m=1}^{10}m^2 - 1\right] = \frac{1}{4}\left[\frac{10 \cdot 11 \cdot 21}{6} - 1\right] = \frac{1}{4}[385-1] = \frac{384}{4} = 96\). Wait, but option B is 96, not A. Let me recalculate to see if I made an error. Actually verifying: sum = \(\frac{1}{4}(4+9+16+25+36+49+64+81+100) = \frac{384}{4} = 96\). So the answer should be B. However, if the JEE answer is A (71), there might be a different interpretation.