the ratio between the coefficient of x^15 and the independent term of x in the expansion of (x^2+2/x)^15, is

To solve the problem, follow these steps:

1. Identify the general term in the binomial expansion of \( \left( x^2 + \frac{2}{x} \right)^{15} \).

T_k = \binom{15}{k} (x^2)^{15-k} \left( \frac{2}{x} \right)^k

2. Simplify to find the term involving \( x \):

T_k = \binom{15}{k} x^{2(15-k)} \cdot \frac{2^k}{x^k} = \binom{15}{k} 2^k x^{30-3k}

3. Coefficient of \( x^{15} \):

Set the exponent \( 30-3k=15 \) to solve for \( k \):

30 - 3k = 15 \implies k = 5

Coefficient is:

\binom{15}{5}2^5=\frac{15!}{5! \cdot10!}\cdot32=96096

4. Independent term, constant term (no \( x \)):

Set the exponent \( 30-3k=0 \) to solve for \( k \):

30 - 3k = 0 \implies k = 10

Coefficient is:

\binom{15}{10}2^{10}=\frac{15!}{10! \cdot5!}\cdot1024=3075072

6. Ratio of coefficients:

\frac{96096}{3075072}=\frac{1}{32}

Thus, the ratio is:

\boxed{\frac{1}{32}}