MCQ
$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = n + 1}^{2n} {\log \left( {1 + \frac{r}{n}} \right) = ..........} $
  • $\log \left( {\frac{{27}}{{4e}}} \right)$
  • B
    $\log \left( {\frac{4}{e}} \right)$
  • C
    $\log \left( {\frac{{25}}{{3{e^2}}}} \right)$
  • D
    $\log \left( {\frac{{27}}{{{e^2}}}} \right)$

Answer

Correct option: A.
$\log \left( {\frac{{27}}{{4e}}} \right)$
$\lim_{x \rightarrow \infty} \frac {1}{n}\sum_{r=n+1}^{2n}\log \left( 1+ \frac {r}{n}\right)$
અહી સંકલનની અધ:સીમા
$= \lim_{x \rightarrow \infty} \frac {r}{n}= \lim_{x \rightarrow \infty} \frac {n+1}{n}$
$= \lim_{x \rightarrow \infty} \left( 1+\frac {1}{n}=1\right)$
સંકલનની ઉર્ધ્વ:સીમા
$= \lim_{x \rightarrow \infty} \frac{r}{n}= \lim_{x \rightarrow \infty} \frac {2n}{n} =2$
$= \lim_{x \rightarrow \infty} \frac{1}{n}= \sum_{r=n+1}^{2n}\log \left(1+ \frac {r}{n}\right)$
$\int^{1}_{0} \log (1+x)dx= [x \log (1+x)]^2_1- \int^{2}_{1} \frac {xdx}{1+x}$
$=(2log3 - log2) - \int^{2}_{1} \frac {xdx}{1+x}$
$= \log \frac {9}{2} -\int^{2}_{1}dx+ \int^{2}_{1} \frac {dx}{1+x}dx$
$= \log \frac {9}{2} -[x]^2_1 + [\log (1+x)]^2_1$
$= \log \frac {9}{2} -[2-1] + [\log(3)-\log(2)]$
$= \log \frac {9}{2}-(1) + \log \frac {3}{2}$
$= \log \frac {27}{4}-1$

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