Question

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

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

Answer 1

Answer: (B), (D)

$\sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}\frac{1}{x+1}dx<\sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}\frac{k+1}{x(x+1)}dx<\sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}\frac{dx}{x}$
$\sum _{k=1}^{98}(\ell n(k+2)-\ln (k+1))<I<\sum _{k=1}^{98}(\ell n(k+1)-\ell nk)$
$\ell n50<I<\ell n99$
$\Rightarrow \frac{49}{50}<I<\ell n99$