Question

If $I=\sum _{k=1}^{98}{\int }_k^{k+1}\frac{k+1}{x(x+1)}dx$, then

• A. $I<\frac{49}{50}$
• B. $I>{\log }_{e}99$
• C. $I>\frac{49}{50}$
• D. $I<{\log }_{e}99$

Answer 1

Answer: (C), (D)

The correct options are
C $I>\frac{49}{50}$
D $I<{\log }_{e}99$

$I={\sum }_{k=1}^{98}{\int }_k^{k+1}\frac{k+1}{x(x+1)}dx$,

$k<x<k+1$

$\therefore \frac{k+1}{x}>1and\frac{k+1}{x+1}<1$

$\therefore \frac{1}{x+1}<\frac{k+1}{x(x+1)}<\frac{1}{x}$

$\therefore {\int }_k^{k+1}\frac{1}{x+1}<{\int }_k^{k+1}\frac{k+1}{x(x+1)}<{\int }_k^{k+1}\frac{1}{x}$

$\therefore ln\left(\frac{k+2}{k+1}\right)<{\int }_k^{k+1}\frac{k+1}{x(x+1)}<ln\left(\frac{k+1}{k}\right)$

$\therefore {\sum }_{k=1}^{98}ln\left(\frac{k+2}{k+1}\right)<{\sum }_{k=1}^{98}{\int }_k^{k+1}\frac{k+1}{x(x+1)}<{\sum }_{k=1}^{98}ln\left(\frac{k+1}{k}\right)$

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

$\therefore ln50<I<ln99$

$ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-....$

$-ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+....$

$\therefore -ln(1-\frac{49}{50})=(\frac{49}{50}+...)$

$\therefore ln50=(\frac{49}{50}+positiveterms)$

$\therefore ln50>\frac{49}{50}$

$\therefore \frac{49}{50}<I<ln99$

So, the answer is option (C) and (D).