Question

If I = $\sum^{98}_{k 1}\int^{k + 1}_k \frac{k + 1}{x(x + 1)} dx$, then
(A) 1 > $log_e$ 99
(B) 1 < $log_e$ 99
(C) 1 < $\frac{49}{50}$
(D) 1 > $\frac{49}{50}$

Answer 1

$\huge \fbox \blue{a} \fbox \purple{n} \fbox \green{s} \fbox \red{w} \fbox \pink{e} \fbox \orange{r}$

$\sf\underline\blue{Please \: note :-}\sf{This \: question \: has\: }$

$\sf {multiple \: correct \: options!}$

$\sf{correct \: option \: is \: (d) I > \frac{49}{50}}⠀\sf{(a)\: I> log_{e} \: 99}$

$\sf\longrightarrow\LARGE\underline\red{Explanation:-}$

$\sf ∫^{k + 1 }_{k} \: \sf\blue{\frac{k + 1}{x(x + 1)}}\sf \: {dx}$

$\sf = (k + 1) ∫^{k + 1}_{k} \: \frac{(x + 1) - x}{x(x + 1)} \: dx$

$\sf = (k + 1) ∫^{k + 1}_{k} \:( \frac{1}{x} - \frac{1}{x + 1})dx$

$\sf =(k+1)[ln \: x \: − \: ln(x+1)] ^{ k+1}$

$\sf =(k+1)([ln(k+1)−ln(k+2)]^{k} −$

$\sf[ln \: k \: − \: ln(k+1)])$

$\sf =(k+1)(2ln(k+1) \: − \: ln(k(k+2))$

$\sf =(k+1)ln \:\left(\frac{(k \: + \: 1)^{2} }{k(k \: + \: 2)}\right)$

$\sf{I \: =}\sf\red{\sum^{98}_{k = 1}}\sf{\: ∫ ^{k + 1}_{k}}\sf\blue{\frac{k \: + \: 1}{x(x \: + \: 1)}}\sf{\:dx}$

$\sf\red{\sum^{98}_{k = 1}}{\: (k + 1) \: In \: \left( \frac{k \: + \: 1}{k \: + \: 2}\right) - k \: In\left(\frac{k}{ k \: + \: 1}\right) -}$

$\sf\red{\sum^{98}_{k = 1}}\sf{In \: k \: - In \: (k \: + \: 1)}$

$\sf = \left(99 \ln \dfrac{99}{100} - \ln \dfrac{1}{2} \right) - (\ln 1 - \ln 99)$

$\sf =ln \: 99 \: − \: ln \: \frac{1}{2}$

$\sf {I = In \: 198}$

$\therefore\sf (d) I > \frac{49}{50}⠀ \sf (a) I > \: log_{e}99 ⠀ ⠀\sf\underline\blue{is\: the \:answer!}$

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$\huge\bold\red{Hope \:it \:helps!}$