The value of I = ∑ _ k = 1 ^ 98 ∫ _ k ^ k + 1 (k + 1)/(x ( x + 1 ))...

$I = \mathop{∑}_{k = 1}^{98}\left(\mathop{∫}_{k}^{k + 1}\dfrac{\left(k + 1\right)}{x\left(x + 1\right)}dx\right)$

$= \mathop{∑}_{k = 1}^{98}\left(k + 1\right)\left(\mathop{∫}_{k}^{k + 1}\left(\dfrac{1}{x} - \dfrac{1}{x + 1}\right)dx\right)$

$= \mathop{∑}_{k = 1}^{98}\left(k + 1\right)\left({\left(ℓnx - ℓn\left(x + 1\right)\right)}_{k}^{k + 1}\right)$

$= \mathop{∑}_{k = 1}^{98}\left(k + 1\right)\left(ℓn\left(k + 1\right) - ℓn\left(k + 2\right) - ℓnk + ℓn\left(k + 1\right)\right)$

$= \mathop{∑}_{k = 1}^{98}\left(\left(k + 1\right)ℓn\left(k + 1\right) - k . ℓnk\right) -\mathop{∑}_{k = 1}^{98}\left(\left(k + 1\right) . ℓn\left(k + 2\right) - k . ℓn\left(k + 1\right)\right) + \mathop{∑}_{k = 1}^{98}\left(ℓn\left(k + 1\right) - ℓnk\right)$ (Difference series)

$∴ I = \left(99ℓn99\right) + \left(- 99ℓn100 + ℓn2\right) + \left(ℓn99\right) = ℓn\left(\dfrac{2 \times {\left(99\right)}^{100}}{{\left(100\right)}^{99}}\right)$ ......(i)

For option (ii):

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

$= {}^{99}{C}_{0} + {}^{99}{C}_{1}\left(99\right) + {}^{99}{C}_{2}{\left(99\right)}^{2} + \ldots + {}^{99}{C}_{97}{\left(99\right)}^{97} + \mathop{\mathop{{}^{99}{C}_{98}{\left(99\right)}^{98}}_{⏟}}_{\left(value = {\left(99\right)}^{99}\right)} + \mathop{\mathop{{}^{99}{C}_{99}{\left(99\right)}^{99}}_{⏟}}_{\left(value = {\left(99\right)}^{99}\right)}$

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

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

$⇒ I < ℓn99$, Hence option (ii) is correct.

For option (iii):

Since, $\mathop{∑}_{k = 1}^{98}\mathop{∫}_{k}^{k + 1}\dfrac{k + 1}{{\left(x + 1\right)}^{2}}dx < \mathop{∑}_{k = 1}^{98}\mathop{∫}_{k}^{k + 1}\dfrac{\left(k + 1\right)dx}{x\left(x + 1\right)}$

For integration L.H.S., we get

$⇒ \mathop{∑}_{k = 1}^{98}\left(\dfrac{1}{k + 2}\right) < I$

$⇒ \mathop{\mathop{\left(\dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \ldots + \dfrac{1}{100}\right)}_{⏟}}_{98terms} < I$

$⇒ \dfrac{98}{100} < \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \ldots + \dfrac{1}{100} < I$

$∴ I > \dfrac{49}{50}$

Hence option (iii) is correct.