Question

The sum to $2n$ terms of the series ${\log }_{3}1-{\log }_{3}3+{\log }_{3}9-{\log }_3\mathrm{27....}$ is

• A. $\frac{(2n-1)}{(n-1)}$
• B. $\frac{(n-1)}{(2n-1)}$
• C. $1-n$
• D. $-n$

Answer 1

Answer: (D)

$-n$

$S={\log }_{3}1-{\log }_{3}3+{\log }_{3}9-{\log }_{3}27+.......-{\log }_3{3}^{2n-1}$

$={\log }_{3}1-{\log }_3{3}^1+{\log }_3{3}^2-{\log }_3{3}^3+.......-{\log }_3{3}^{2n-1}$

$=0-1+2-3+4-5+.......-(2n-1)$

$=-(1+3+5+.......2n-1)+(2+4+.......(2n-2\left)\right)$

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

$\because [\text{Sum of n terms of an A.P}=\frac{n}{2}[\text{First term}+\text{Last term}\left]\right]$

$\Rightarrow S=\frac{-n}{2}\times 2n+\frac{n-1}{2}\times 2n$

$=-n^2+(n-1)\left(n\right)$

$\Rightarrow S=-n$