Question

The sum of series $\sum _{x=1}^n\log \left(\frac{2^x}{3^{x-1}}\right)$ is

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

Answer 1

The correct option is B

$\log {\left(\frac{2^{n+1}}{3^{n-1}}\right)}^{\frac{n}{2}}$

Explanation for the correct answer:

The given series is $\sum _{x=1}^n\log \left(\frac{2^x}{3^{x-1}}\right)$

The series can be written in a simplified format by substituting the values of of $x$ from $1$through $n$

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

$=\log \left(\frac{2^1}{1}\times \frac{2^2}{3^1}\times \frac{2^3}{3^2}\times ....\times \frac{2^n}{3^{n-1}}\right)$ $\left[\because \log \left(a\right)+\log \left(b\right)+\log \left(c\right)=\log \left(abc\right)\right]$

$=\log \left(\frac{2^1\times 2^2\times 2^3\times ....\times 2^n}{3^1\times 3^2\times ....\times 3^{n-1}}\right)$ $\left[\because 1+2+3+...+n=\frac{n\left(n+1\right)}{2}\right]$

$=\log \left(\frac{2^{{\displaystyle \frac{n\left(n+1\right)}{2}}}}{3^{{\displaystyle \frac{n\left(n-1\right)}{2}}}}\right)$ $\left[\because \frac{a^{xy}}{b^{xz}}={\left(\frac{a^y}{b^z}\right)}^x\right]$

$=\log {\left(\frac{2^{n+1}}{3^{n-1}}\right)}^{\frac{n}{2}}$

Hence the sum of the series $\sum _{x=1}^n\log \left(\frac{2^x}{3^{x-1}}\right)$ is $\log {\left(\frac{2^{n+1}}{3^{n-1}}\right)}^{\frac{n}{2}}$.

Hence, option C is the correct answer.