Question

The value of ${\sum }_{n=1}^m\log \frac{a^{2n-1}}{b^{m-1}}(a\ne 0;b\ne 0,1)$ is equal to

• A. $m\log \frac{a^{2m}}{b^{m-1}}$
• B. $\log \frac{a^{2m}}{b^{m-1}}$
• C. $\frac{m}{2}\log \frac{a^{2m}}{b^{2m-2}}$
• D. $\frac{m}{2}\log \frac{a^{2m}}{b^{m+1}}$

Answer 1

The correct option is C $\frac{m}{2}\log \frac{a^{2m}}{b^{2m-2}}$

Let $S={\sum }_{n=1}^m\log \frac{a^{2n-1}}{b^{m-1}}$
$\Rightarrow S={\sum }_{n=1}^m(\log a^{2n-1}-\log b^{m-1})=\log a{\sum }_{n=1}^m(2n-1)-(m-1)\log b{\sum }_{n=1}^m$
$\Rightarrow S=-m(m-1)\log b+\log a\{2{\sum }_{n=1}^{m}n-{\sum }_{n=1}^m\}$
$\Rightarrow S=-m(m-1)\log b+\log a[m(m+1)-m]$
$\Rightarrow S=\frac{m}{2}[2m\log a-(2m-2)\log b]$
$\Rightarrow S=\frac{m}{2}\log \frac{a^{2m}}{b^{2m-2}}$
Ans: C