Question

$\underset{n\to \infty }{lim}{\left(0.2\right)}^{{\log }_{\sqrt{5}}(1/4+1/8+1/16+...tonterms)}$ is equal to

• A. $2$
• B. $4$
• C. $8$
• D. $0$

Answer 1

Answer: (B)

$4$

$\underset{n\to \infty }{lim}{\left(\frac{1}{5}\right)}^{{\log }_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...\infty )}=\underset{n\to \infty }{lim}{5}^{{}^{-2{\log }_5\left(\frac{1}{4}\frac{1-{\left(\frac{1}{2}\right)}^n}{1-\frac{1}{2}}\right)}}=\underset{n\to \infty }{lim}{5}^{{}^{{\log }_5{\left(\frac{1}{4}\frac{1-{\left(\frac{1}{2}\right)}^n}{1-\frac{1}{2}}\right)}^{-2}}}=\underset{n\to \infty }{lim}{\left(\frac{1}{4}\frac{1-{\left(\frac{1}{2}\right)}^n}{1-\frac{1}{2}}\right)}^{-2}={(\frac{1}{4}\times \frac{2}{1})}^{-2}=4$