Power of JEE 2017 advanced

$\begin{aligned} I & = \sum_{k=1}^{98} \int_k^{k+1} \frac {k+1}{x(x+1)} dx \\ & = \sum_{k=1}^{98} (k+1) \int_k^{k+1} \left(\frac 1x - \frac 1{x+1} \right) dx \\ & = \sum_{k=1}^{98} (k+1) \bigg[ \ln (x) - \ln (x+1) \bigg]_k^{k+1} \\ & = \sum_{k=1}^{98} (k+1) \left(\ln \left(\frac {k+1}{k+2}\right) - \ln \left(\frac k{k+1}\right)\right) \\ & = \sum_{k=1}^{98} \left((k+1) \ln \left(\frac {k+1}{k+2}\right) - k \ln \left(\frac k{k+1}\right) - \ln \left(\frac k{k+1}\right) \right) \\ & = \sum_{\color{#3D99F6}k=1}^{\color{#3D99F6}98} (k+1) \ln \left(\frac {k+1}{k+2}\right) - \sum_{k=1}^{98} k \ln \left(\frac k{k+1}\right) - \sum_{k=1}^{98} \ln k + \sum_{\color{#3D99F6}k=1}^{\color{#3D99F6}98} \ln (k+1) \\ & = \sum_{\color{#D61F06}k=2}^{\color{#D61F06}99} k \ln \left(\frac k{k+1}\right) - \sum_{k=1}^{98} k \ln \left(\frac k{k+1}\right) - \sum_{k=1}^{98} \ln k + \sum_{\color{#D61F06}k=2}^{\color{#D61F06}99} \ln k \\ & = 99 \ln \left(\frac {99}{100}\right) - \ln \left(\frac 12 \right) - \ln 1 + \ln 99 \\ & = \ln 99 - 99 \ln \left(\frac {100}{99} \right) + \ln 2 \end{aligned}$

Then we have

$\begin{aligned} I & = \ln 198 - \ln \left(1+\frac 1{99}\right)^{99} > \log_3 81 - \ln e = 3 > \frac {49}{50} \end{aligned}$

And

$\begin{aligned} I & = \ln 99 - 99 \ln \left(\frac {100}{99} \right) + \ln 2 = \ln 99 - 99 \ln 100 + 99 \ln 99 + 99 \ln 2^{\frac 1{99}} \\ & = \ln 99 + 99 \ln \frac {99 \times 2^{\frac 1{99}}}{100} < \ln 99 + 99 \ln \frac {100 \times 2^{\frac 1{99}}}{100} = \ln 99 + \frac {\ln 2}{99} < \ln 99 \end{aligned}$

Therefore, $\implies \boxed{\frac {49}{50} < I < \ln 99}$ .

I cheated on the exam for this. I used induction. Instead of keeping the upper limit of the sum to be 98 I kept it as 2 for the initial step. And then assumed the bounds to exist suitably for the given question. For example for 2 I kept the upper and lower limits as $ln(2+1)$ and $2/(2+2)$ . Why I wrote those that way will be clear immediately. For $n$ I kept the bounds as $ln(n+1)$ and $n/(n+2)$ and henceforth the induction becomes trivial.