QJEEAdv17Ma66f1: JEE Advanced 2017 Mathematics - Class 12 JEEAdv17Ma66f1 Solution: If , then

Question

Verified Solution • 2025–26 Session moderate

If

I = \sum\nolimits_{k = 1}^{98} {\int_k^{k + 1} {{{k + 1} \over {x(x + 1)}}} dx}

, then

Accepted Answer

I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x(x + 1)}}dx} }

Clearly,

I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x{{(x + 1)}^2}}}dx} }

\Rightarrow I &gt; \sum\limits_{k = 1}^{98} {(k + 1)\int_k^{k + 1} {{1 \over {{{(x + 1)}^2}}}dx} }

\Rightarrow I &gt; \sum\limits_{k = 1}^{98} {( - (k + 1))\left[ {{1 \over {k + 2}} - {1 \over {k + 1}}} \right]}

\Rightarrow I &gt; \sum\limits_{k = 1}^{98} {{1 \over {k + 2}}}

\Rightarrow I &gt; {1 \over 3} + ... + {1 \over {100}} &gt; {{98} \over {100}}

\Rightarrow I &gt; {{49} \over {50}}

Also,

I = \sum\limits_{k = 1}^{98} {\int_k^{k + 1} {{{(k + 1)} \over {x(x + 1)}}dx} }

= \sum\limits_{k = 1}^{98} {[{{\log }_e}(k + 1) - {{\log }_e}k]}

I &lt; {\log _e}99

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