Question

$(1+3^{-1})(1+3^{-2})(1+3^{-4})(1+3^{-8})\dots ..(1+3^{-2n})$ is equal to

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

Answer 1

The correct option is C

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

Explanation for the correct option:

Let the given series be, $S=(1+3^{-1})(1+3^{-2})(1+3^{-4})...(1+3^{-2n})$

Multiplying both sides by $\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{3}}^{\mathbf{-}\mathbf{1}}\mathbf{)}$,

$\Rightarrow$$(1-3^{-1})S=(1-3^{-1})(1+3^{-1})(1+3^{-2})(1+3^{-4})...(1+3^{-2n})$

$\Rightarrow$ $\frac{2}{3}S=(1-3^{-2})(1+3^{-2})(1+3^{-4})...(1+3^{-2n})$

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

$\Rightarrow$ $\frac{2}{3}S=\left(1-3^{-2n}\right)\left(1+3^{-2n}\right)$ $\left[\mathbf{\because }{\mathit{a}}^{\mathbf{2}}\mathbf{-}{\mathit{b}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\left(a-b\right)\left(a+b\right)\right]$

$\Rightarrow$ $\frac{2}{3}S=(1-3^{-2n+1})$

$\Rightarrow$ $S=\frac{\mathbf{3}}{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{3}}^{\mathbf{-}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{)}$

Hence, Option ‘C’ is Correct.