i) Step 1: Write down the binomial expansion formula. The binomial expansion for $(1+y)^n$ is given by: $$(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots$$ In this problem, $y = \frac{1}{3}x$ and $n=8$. Step 2: Expand the given expression up to the $x^2$ term. Substitute $y = \frac{1}{3}x$ and $n=8$ into the formula: $$\left(1+\frac{1}{3}x\right)^8 = 1 + 8\left(\frac{1}{3}x\right) + \frac{8(8-1)}{2!}\left(\frac{1}{3}x\right)^2 + \dots$$ $$\left(1+\frac{1}{3}x\right)^8 = 1 + \frac{8}{3}x + \frac{8 \times 7}{2}\left(\frac{1}{9}x^2\right) + \dots$$ $$\left(1+\frac{1}{3}x\right)^8 = 1 + \frac{8}{3}x + 28\left(\frac{1}{9}x^2\right) + \dots$$ $$\left(1+\frac{1}{3}x\right)^8 = 1 + \frac{8}{3}x + \frac{28}{9}x^2 + \dots$$ Step 3: Compare the expanded form with the given expression to find $a$ and $b$. Given: $\left(1+\frac{1}{3}x\right)^8 = 1 + \frac{2}{3}ax - \frac{4}{9}bx^2 + \dots$ Comparing the coefficient of $x$: $$\frac{8}{3}x = \frac{2}{3}ax$$ $$8 = 2a$$ $$a = \frac{8}{2}$$ $$a = 4$$ Comparing the coefficient of $x^2$: $$\frac{28}{9}x^2 = -\frac{4}{9}bx^2$$ $$28 = -4b$$ $$b = \frac{28}{-4}$$ $$b = -7$$ The values of the constants are $\boxed{a=4, b=-7}$. ii) Step 1: Understand the problem. We are given the sum of the first $n$ terms of a sequence, $S_n = \frac{n}{2}(3n+1)$. We need to find the sum of the terms from the $10^{th}$ term to the $30^{th}$ term. This can be calculated as $S_{30} - S_9$. Step 2: Calculate $S_{30}$. Substitute $n=30$ into the formula for $S_n$: $$S_{30} = \frac{30}{2}(3(30)+1)$$ $$S_{30} = 15(90+1)$$ $$S_{30} = 15(91)$$ $$S_{30} = 1365$$ Step 3: Calculate $S_9$. Substitute $n=9$ into the formula for $S_n$: $$S_9 = \frac{9}{2}(3(9)+1)$$ $$S_9 = \frac{9}{2}(27+1)$$ $$S_9 = \frac{9}{2}(28)$$ $$S_9 = 9 \times 14$$ $$S_9 = 126$$ Step 4: Calculate the sum from the $10^{th}$ term to the $30^{th}$ term. Sum = $S_{30} - S_9$ Sum = $1365 - 126$ Sum = $1239$ The sum of the terms from the $10^{th}$ term to the $30^{th}$ term is $\boxed{1239}$.