To find the radius of convergence for the given power series, we need to analyze the series:

$\sum_{n=1}^\infty \frac{5^n \cdot n! \cdot (4n+2)! \cdot (2n)!!}{6^n \cdot [(n+3)!]^4 \cdot (4n-1)!!} (-3x + 4)^n$

The radius of convergence $R$ can be determined using the ratio test. First, let's simplify the expression inside the summation:

1. Double factorial expressions:

   • For an odd $n$: $(2n-1)!! = (2n-1) \times (2n-3) \times \cdots \times 3 \times 1$
   • For an even $n$: $(2n)!! = 2^n \cdot n!$
   • For $(4n-1)!!$ and $(2n)!!$, we need to apply these definitions accordingly.

2. Apply the ratio test: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$ where $a_n = \frac{5^n \cdot n! \cdot (4n+2)! \cdot (2n)!!}{6^n \cdot [(n+3)!]^4 \cdot (4n-1)!!} (-3x + 4)^n$.

To simplify, let's compute the ratio $\left| \frac{a_{n+1}}{a_n} \right|$:

[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{5^{n+1} \cdot (n+1)! \cdot (4(n+1)+2)! \cdot (2(n+1))!!}{6^{n+1} \cdot [(n+4)!]^4 \cdot (4(n+1)-1)!!} \cdot \frac{6^n \cdot [(n+3)!]^4 \cdot (4n-1)!!}{5^n \cdot n! \cdot (4n+2)! \cdot (2n)!!} (-3x+4) \right| ]

Simplify the ratio:

$= \left| \frac{5 \cdot 5^n \cdot (n+1) \cdot n! \cdot (4n+6)! \cdot 2 \cdot (2n)!!}{6 \cdot 6^n \cdot (n+1+3)! \cdot (n+3)!^3 \cdot (4n+3)!!} \cdot \frac{6^n \cdot (n+3)!^4 \cdot (4n-1)!!}{5^n \cdot n! \cdot (4n+2)! \cdot (2n)!!} (-3x+4) \right|$

$= \left| \frac{5 \cdot 2 \cdot (4n+6)(4n+5)(4n+4)! \cdot (2n+2)(2n)!}{6 \cdot 6^n \cdot 6 \cdot (n+3)!^3 \cdot (4n+3)(4n+2)(4n+1)(4n)!} \cdot \frac{6^n \cdot (4n-1)!!}{(4n+2)! \cdot (2n)!!} (-3x+4) \right|$

Simplify the factorial terms and constants:

$= \left| \frac{10 \cdot 5^n \cdot (4n+6)(4n+5)(4n+4)! \cdot (2n+2)}{6^{n+1} \cdot (4n+3)(4n+2)(4n+1)(4n)!} \cdot (-3x+4) \right|$

Using Stirling's approximation for large $n$:

$n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$

Applying this to the factorials:

$\lim_{n \to \infty} \left| \frac{5}{6} \cdot \frac{5^n}{6^n} \cdot \frac{10 \cdot (4n+6)(4n+5)}{(4n+3)(4n+2)(4n+1)} (-3x+4) \right|$

This simplifies further, but the dominant terms are:

$\left| \frac{5}{6} (-3x+4) \right| \to 0$

Thus, the radius of convergence $R$ is given by the reciprocal of the limit of the ratio:

$R = \frac{1}{\lim_{n \to \infty} \left| \frac{5}{6} \right|}$

Finally,

$R = \frac{6}{5}$

So, the radius of convergence $R$ is $\frac{6}{5}$.

Would you like more details or have any questions? Here are five more questions you might find interesting:

1. What is the radius of convergence for the series $\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} x^n$?
2. How do you find the interval of convergence for a power series?
3. Can you explain the ratio test in more detail?
4. What is the relationship between the radius of convergence and the interval of convergence?
5. How do you apply Stirling's approximation to factorials in series convergence problems?

Tip: When using the ratio test, always ensure to simplify the terms carefully to avoid mistakes in calculating the limit.