Question

The sum of $n$ terms of the series ${\sum }_{x=1}^n\log \frac{2^x}{3^{x-1}}$ is

• A. $\log {\left(\frac{2^{n-1}}{3^{n+1}}\right)}^{n/2}$
• B. $\log {\left(\frac{2^{n+1}}{3^{n-1}}\right)}^{n/2}$
• C. $\log {\left(\frac{3^{n-1}}{2^{n+1}}\right)}^{n/2}$
• D. $\log {\left(\frac{3^{n+1}}{2^{n-1}}\right)}^{n/2}$

Answer 1

The correct option is B $\log {\left(\frac{2^{n+1}}{3^{n-1}}\right)}^{n/2}$

$\therefore {\sum }_{x=1}^n\frac{2^x}{3^{x-1}}=\log \left(\frac{2^1}{3^0}\right)+\log \left(\frac{2^2}{3^1}\right)+\log \left(\frac{2^3}{3^2}\right)+....+\log \left(\frac{2^n}{3^{n-1}}\right)$
$=\log \left(\frac{2^1\times 2^2\times 2^3\times ....\times 2^n}{3^1\times 3^2\times ....\times 3^{n-1}}\right)=\log {\left(\frac{2^{n+1}}{3^{n-1}}\right)}^{n/2}$