Question

The value of $\sum _{r=1}^n\log \left(\frac{a^r}{b^{r-1}}\right)$ is

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

Answer 1

The correct option is C

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

The explanation for the correct option

The given expression: $\sum _{r=1}^n\log \left(\frac{a^r}{b^{r-1}}\right)$.

$$\begin{gathered} \sum _{r=1}^n\log \left(\frac{a^r}{b^{r-1}}\right)=\log \left(\frac{a^1}{b^0}\right)+\log \left(\frac{a^2}{b^1}\right)+....+\log \left(\frac{a^n}{b^{n-1}}\right) \\ \Rightarrow \sum _{r=1}^n\log \left(\frac{a^r}{b^{r-1}}\right)=\log \left(\frac{a^1}{b^0}\times \frac{a^2}{b^1}\times ......\times \frac{a^n}{b^{n-1}}\right)\left[\because \log \left(AB\right)=\log \left(A\right)+\log \left(B\right)\right] \\ \Rightarrow \sum _{r=1}^n\log \left(\frac{a^r}{b^{r-1}}\right)=\log \left(\frac{a^{1+2+....+n}}{b^{1+2+....+\left(n-1\right)}}\right) \end{gathered}$$

Now, the sum of the first $n$ natural can be given by, $S_n=\frac{n\left(n+1\right)}{2}$.

we know that $1+2+.....+n=\frac{n\left(n+1\right)}{2}$.

And, $1+2+.....+\left(n-1\right)=\frac{\left(n-1\right)\left(n-1+1\right)}{2}=\frac{n\left(n-1\right)}{2}$.

Thus, $\sum _{r=1}^n\log \left(\frac{a^r}{b^{r-1}}\right)=\log \left(\frac{a^{1+2+....+n}}{b^{1+2+....+\left(n-1\right)}}\right)$

$$\begin{gathered} \Rightarrow \sum _{r=1}^n\log \left(\frac{a^r}{b^{r-1}}\right)=\log \left(\frac{a^{{\displaystyle \frac{n\left(n+1\right)}{2}}}}{b^{{\displaystyle \frac{n\left(n-1\right)}{2}}}}\right) \\ \Rightarrow \sum _{r=1}^n\log \left(\frac{a^r}{b^{r-1}}\right)=\log {\left(\frac{a^{{\displaystyle n+1}}}{b^{{\displaystyle n-1}}}\right)}^{\frac{n}{2}} \\ \Rightarrow \sum _{r=1}^n\log \left(\frac{a^r}{b^{r-1}}\right)=\frac{n}{2}\log \left(\frac{a^{{\displaystyle n+1}}}{b^{{\displaystyle n-1}}}\right)\left[\because \log {\left(A\right)}^a=a\log \left(A\right)\right] \end{gathered}$$

Therefore, the value of $\sum _{r=1}^n\log \left(\frac{a^r}{b^{r-1}}\right)$ is $\frac{n}{2}\log \left(\frac{a^{n+1}}{b^{n-1}}\right)$.

Hence, the correct option is (C).