Question

If $L={\sum }_{r=7}^{2400}{\log }_7\left(\frac{r+1}{r}\right),M={\prod }_{r=2}^{1023}{\log }_r(r+1)$ and $N={\sum }_{r=2}^{2011}\left(\frac{1}{{\log }_{r}p}\right)$ where p=$(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...........\cdot 2011)$, then

• A. $L+M=13$
• B. $M^2+N^2=101$
• C. $L-M+N=6$
• D. $LMN=30$

Answer 1

The correct option is D $LMN=30$

$\text{Solve:}$

$\text{Given,}$

$L={\sum }_{r=7}^{2400}{\log }_7\left(\frac{r+1}{r}\right)$

$\Rightarrow L={\log }_7\left(\frac{8}{7}\right)+{\log }_7\left(\frac{9}{8}\right)+{\log }_7\left(\frac{10}{9}\right)+-\cdots +{\log }_7\left(\frac{2401}{2400}\right)$

$\text{We know that}$

$\log a+\log b=\log ab$

$\Rightarrow L={\log }_7\left(\frac{8}{7}\right)\times \left(\frac{9}{8}\right)\times \left(\frac{10}{9}\right)\times \dots \times \left(\frac{401}{2400}\right)$

$\Rightarrow L={\log }_7\left(\frac{240}{7}\right)={\log }_7{7}^3$

$\Rightarrow L=3-\left(i\right)\{{\log }_a{a}^x=x$

$M={\pi }_{r=2}^{1023}{\log }_r(r+1)$

$\Rightarrow M={\log }_2\left(3\right)\cdot {\log }_{3}4\cdot {\log }_{4}5..{\log }_{1023}1024$

$\Rightarrow M=\left(\frac{\log 3}{\log 2}\right)\cdot \left(\frac{\log 4}{\log 3}\right).\cdots \left(\frac{\log 1024}{\log 1023}\right)$

$\{{\log }_{a}b=\frac{\log b}{\log a}\}$

$\Rightarrow M=\frac{\log 1024}{\log 2}={\log }_2{2}^{10}$

$\Rightarrow M=10-\left(ii\right)$

Now, $N={\sum }_{r=2}^{20\pi }\left(\frac{1}{{\log }_{r}p}\right)$

Where, $P=(1.2.3.4.5\dots ..2011)$

$\Rightarrow P=2011!$

$\Rightarrow N={\log }_p^2+{\log }_p^3+{\log }_p^4+\dots +{\log }_p^{2011}$

$\Rightarrow N={\log }_p(2\cdot 3\cdot 4\cdot \dots 2011)={\log }_p\left(2011\right)!$

$\Rightarrow N={\log }_p^p=1-\left(iii\right)$

on adding equation (i) and equation (ii)

we get.

$L+M=3+10=13$

from eqn. (i), (i) and (iii) we get

$L=3,M=10$ and $N=1$

$\Rightarrow M^2+N^2=100+1=101$

and LMN $=3\cdot 10\cdot 1=30$