Answer
(a) $\frac{58025}{59049}$
(b) $2+\sqrt 2$
Work Step by Step
(a) Test for geometric sequence: $\frac{a_2}{a_1}=\frac{2/3^2}{1/3}=\frac{2}{3}$, $\frac{a_3}{a_2}=\frac{2^2/3^3}{2/3^2}=\frac{2}{3}$, ... $\frac{a_{10}}{a_9}=\frac{2^9/3^{10}}{2^8/3^9}=\frac{2}{3}$. Thus it is a geometric sequence with $a_1=\frac{1}{3}, r=\frac{2}{3}, n=10$. Use the formula for the sum, we have:
$Sum_1=a_1\frac{1-r^n}{1-r}=\frac{1}{3}\times\frac{1-(2/3)^{10}}{1-2/3}=\frac{3^{10}-2^{10}}{3^{10}}=\frac{58025}{59049}$
(b) Test for geometric sequence: $\frac{b_2}{b_1}=\frac{1/2^{1/2}}{1}=\frac{1}{\sqrt 2}$, $\frac{b_3}{b_2}=\frac{1/2}{1/2^{1/2}}=\frac{1}{\sqrt 2}$, ... Thus it is a geometric sequence with $b_1=1, r=\frac{1}{\sqrt 2}$. Use the formula for the infinite sum, we have:
$Sum_2=\frac{b_1}{1-r}=\frac{1}{1-\frac{1}{\sqrt 2}}=\frac{\sqrt 2}{\sqrt 2-1}=\frac{\sqrt 2(\sqrt 2+1)}{(\sqrt 2-1)(\sqrt 2+1)}=2+\sqrt 2$
