The value of I = ∑ k = 1 98 ∫ k k + 1 k + 1 x ( x + 1 ) d x ,   then

$I= \sum _{k=1}^{98}\left(\underset{k}{\overset{k+1}{\int }}\frac{\left(k+1\right)}{x\left(x+1\right)}dx\right)$

$= \sum _{k=1}^{98}\left(k+1\right)\left(\underset{k}{\overset{k+1}{\int }}\left(\frac{1}{x}-\frac{1}{x+1}\right)dx\right)$

$= \sum _{k=1}^{98}\left(k+1\right)\left({\left(\ell n x-\ell n\left(x+1\right)\right)}_k^{k+1}\right)$

$= \sum _{k=1}^{98}\left(k+1\right) \left(\ell n \left(k+1\right)-\ell n\left(k+2\right)-\ell n k+\ell n\left(k+1\right)\right)$

$= \sum _{k=1}^{98}\left(\left(k+1\right)\ell n \left(k+1\right)-k.\ell n k\right)- \sum _{k=1}^{98}\left(\left(k+1\right).\mathcal{ }\ell n \left(k+2\right)-k.\ell n\left(k+1\right)\right)+\sum _{k=1}^{98}\left(\ell n\left(k+1\right)-\ell n k\right)$    (Difference series)

$\therefore I=\left(99 \ell n 99\right)+\left(-99 \ell n 100+ \ell n 2\right)+\left(\ell n 99\right)=\mathcal{ }\ell n \left(\frac{2\times {\left(99\right)}^{100}}{{\left(100\right)}^{99}}\right)$      ......(i)

For option (ii):

Now, consider  ${\left(100\right)}^{99}={\left(1+99\right)}^{99}$

$= { }^{99}{C}_0+{ }^{99}{C}_1\left(99\right)+{ }^{99}{C}_2{\left(99\right)}^2+\dots +{ }^{99}{C}_{97} {\left(99\right)}^{97}+ \underset{\left(\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}={\left(99\right)}^{99}\right)}{\underbrace{{ }^{99}{C}_{98}{\left(99\right)}^{98}}}+ \underset{\left(\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}={\left(99\right)}^{99}\right)}{\underbrace{{ }^{99}{C}_{99}{\left(99\right)}^{99}}}$

$\Rightarrow {\left(100\right)}^{99}>2. {\left(99\right)}^{99} \Rightarrow \frac{2\times {\left(99\right)}^{99}}{{\left(100\right)}^{99}}<1$

$\therefore \frac{2\times {\left(99\right)}^{100}}{{\left(100\right)}^{99}}<99$    (on multiplying by 99)

$\Rightarrow I<\mathcal{ }\ell n 99$, Hence option (ii) is correct.

For option (iii):

Since,    $\sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}\frac{k+1}{{\left(x+1\right)}^2} dx< \sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}\frac{\left(k+1\right) dx}{x\left(x+1\right)}$

For integration L.H.S., we get

$\Rightarrow \sum _{k=1}^{98}\left(\frac{1}{k+2}\right)<I$

$\Rightarrow \underset{98 \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s}}{\underbrace{\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots +\frac{1}{100}\right)}}<I$

$\Rightarrow \frac{98}{100}<\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots +\frac{1}{100}<I$

$\therefore I>\frac{49}{50}$

Hence option (iii) is correct.