Here are the solutions to the problems. 1. The coefficient ratio of the coefficient of $x^4$ to that of $x^3$ in the binomial expansion of $(1+2x)^n$ is $3:1$. Find the value of $n$. Step 1: Write the general term of the binomial expansion. The general term $T_{k+1}$ for $(1+2x)^n$ is given by: $$T_{k+1} = \binom{n}{k} (1)^{n-k} (2x)^k = \binom{n}{k} 2^k x^k$$ Step 2: Find the coefficient of $x^3$. For $x^3$, we set $k=3$. The coefficient of $x^3$ is $\binom{n}{3} 2^3 = 8\binom{n}{3}$. Step 3: Find the coefficient of $x^4$. For $x^4$, we set $k=4$. The coefficient of $x^4$ is $\binom{n}{4} 2^4 = 16\binom{n}{4}$. Step 4: Set up the ratio and solve for $n$. The ratio of the coefficient of $x^4$ to that of $x^3$ is $3:1$. $$\frac{16\binom{n}{4}}{8\binom{n}{3}} = \frac{3}{1}$$ $$2 \frac{\binom{n}{4}}{\binom{n}{3}} = 3$$ Using the identity $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$, with $k=4$: $$2 \left(\frac{n-4+1}{4}\right) = 3$$ $$2 \left(\frac{n-3}{4}\right) = 3$$ $$\frac{n-3}{2} = 3$$ $$n-3 = 6$$ $$n = 9$$ The value of $n$ is $\boxed{\text{9}}$. 2. The first three terms of the expansion of $(1+ax)^n$ is $1+2x+\frac{15}{8}x^2+...$. Find the value of $a$ and $n$. Step 1: Write the first three terms of the binomial expansion of $(1+ax)^n$. $$(1+ax)^n = \binom{n}{0} (1)^n (ax)^0 + \binom{n}{1} (1)^{n-1} (ax)^1 + \binom{n}{2} (1)^{n-2} (ax)^2 + \dots$$ $$(1+ax)^n = 1 + n