$ \begin{aligned} I & =\sum_{k=1}^{98} \int_k^{k+1} \frac{(k+1)}{x(x+1)} d x \\ I & =\sum_{k=1}^{98}(k+1)\left(\int_k^{k+1}\left(\frac{1}{x}-\frac{1}{x+1}\right) d x\right) \\ & =\sum_{k=1}^{98}(k+1)\left((\ell n x-\ell n(x+1))_k^{k+1}\right. \\ & =\sum_{k=1}^{98}(k+1)((\ell n(k+1)-\ell n(k+2)-\ell n k+\ell n(k+1)) \\ & =\sum_{k=1}^{98}(k+1)(\ell n(k+1)-k \cdot \ell n k)-\sum_{k=1}^{98}((k+1) \cdot \ell n k+2-k \cdot \ell n k+1)+\sum_{k=1}^{98}(\ell n k(k+1)-\ell n k) \end{aligned} $ (Difference series) $ \therefore \mathrm{I}=(99 \ell \mathrm{n} 99)+(-99 \ell \mathrm{n} 100+\ell \mathrm{n} 2)+(\ell \mathrm{n} 99)=\ell \mathrm{n}\left(\frac{2 \times(99)^{100}}{(100)^{99}}\right) $ For option (B) : Now Consider $(100)^{99}=(1+99)^{99}$ $ \begin{aligned} & ={ }^{99} \mathrm{C}_0+{ }^{99} \mathrm{C}_1(99)+{ }^{99} \mathrm{C}_2(99)^2+\ldots . .+{ }^{99} \mathrm{C}_{97}(99)^{97}+\underbrace{99}_{\text {(value=(99)(an) }} \mathrm{C}_{98}(99)^{98} \\ & \Rightarrow 100^{99}>2 .(99)^{99} \Rightarrow \frac{2 \times(99)^{99}}{(100)^{99}}<1 \\ & \therefore \frac{2 \times(99)^{99}(99)^{99}}{(100)^{99}}<99 \text { (on multiplying by } 99 \text { ) } \\ & \Rightarrow \mathrm{I}<\ell \mathrm{n} 99 \end{aligned} $ For option (C) : Since, $\sum_{\mathrm{k}=1}^{98} \int_{\mathrm{k}}^{\mathrm{k}+1} \frac{\mathrm{k}+1}{(\mathrm{x}+1)^2} \mathrm{dx} <\sum_{\mathrm{k}=1}^{98} \int_{\mathrm{k}}^{\mathrm{k}+1} \frac{(\mathrm{k}+1) \mathrm{dx}}{\mathrm{x}(\mathrm{x}+1)}$ $ \Rightarrow \sum_{\mathrm{k}=1}^{98}\left(\frac{1}{\mathrm{k}+2}\right)<\mathrm{I} $ (on integration) $ \begin{aligned} & \Rightarrow \underbrace{\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots \ldots+\frac{1}{100}\right)}_{98 \text { terms }}<\mathrm{I} \\ & \Rightarrow \frac{98}{100}<\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots . .+\frac{1}{100}<\mathrm{I} \\ & \therefore \mathrm{I}>\frac{49}{50} \end{aligned} $ Hence option (C) is correct.