Write the general term of the binomial expansion.

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 $(1+ax)^n$.

$$(1+ax)^n = \binom{n}{0} (1)^n (ax)^0 + \binom{n}{1} (1)^{n-1} (ax)^1 + \binom{n}{2} (1)^{n-2} (ax)^2 + \dots$$ $$$$