Let $\{a_{n}\}$ be a sequence which satisfy

$a_{1}=5$ and $a_{n=}\sqrt[n]{a_{n-1}^{n-1}+2^{n-1}+2.3^{n-1}} \qquad \forall n\geq2$

[b](a)[/b] Find the general fomular for $a_{n}$

[b](b)[/b] Prove that $\{a_{n}\}$ is decreasing sequences
1. **(a) Find the general formula for \( a_n \)**

   Given the sequence \( \{a_n\} \) defined by:
   \[
   a_1 = 5
   \]
   and
   \[
   a_n = \sqrt[n]{a_{n-1}^{n-1} + 2^{n-1} + 2 \cdot 3^{n-1}} \quad \forall n \geq 2
   \]

   Let's define \( v_n = a_n^n \). Then, we have:
   \[
   v_n = a_{n-1}^{n-1} + 2^{n-1} + 2 \cdot 3^{n-1}
   \]

   For \( n = 2 \):
   \[
   v_2 = a_1^1 + 2^1 + 2 \cdot 3^1 = 5 + 2 + 6 = 13
   \]

   For \( n = 3 \):
   \[
   v_3 = a_2^2 + 2^2 + 2 \cdot 3^2 = 13 + 4 + 18 = 35
   \]

   We observe that:
   \[
   v_n - v_{n-1} = 2^{n-1} + 2 \cdot 3^{n-1}
   \]

   Summing up the differences from \( v_1 \) to \( v_n \):
   \[
   v_n - v_1 = \sum_{k=2}^{n} (2^{k-1} + 2 \cdot 3^{k-1})
   \]

   This simplifies to:
   \[
   v_n - 5 = (2^1 + 2^2 + \cdots + 2^{n-1}) + (2 \cdot 3^1 + 2 \cdot 3^2 + \cdots + 2 \cdot 3^{n-1})
   \]

   Using the sum of geometric series:
   \[
   v_n - 5 = (2^n - 2) + 2(3^n - 3)
   \]

   Therefore:
   \[
   v_n = 2^n + 3^n
   \]

   Since \( v_n = a_n^n \), we have:
   \[
   a_n = \sqrt[n]{2^n + 3^n}
   \]

   Thus, the general formula for \( a_n \) is:
   \[
   \boxed{a_n = \sqrt[n]{2^n + 3^n}}
   \]

2. **(b) Prove that \( \{a_n\} \) is a decreasing sequence**

   We need to show that \( a_n > a_{n+1} \) for all \( n \geq 1 \).

   Given:
   \[
   a_n = \sqrt[n]{2^n + 3^n}
   \]

   Let:
   \[
   f_n = \left( \frac{2}{3} \right)^n
   \]

   Then:
   \[
   a_n = 3 \sqrt[n]{1 + f_n}
   \]

   Since \( 0 < \frac{2}{3} < 1 \), it follows that \( 0 < f_n < f_{n-1} \). Therefore:
   \[
   1 < 1 + f_n < 1 + f_{n-1}
   \]

   Taking the \( n \)-th root:
   \[
   1 < (1 + f_n)^{\frac{1}{n}} < (1 + f_{n-1})^{\frac{1}{n}}
   \]

   Since \( \frac{1}{n} < \frac{1}{n-1} \) for all \( n \geq 2 \), we have:
   \[
   (1 + f_{n-1})^{\frac{1}{n}} < (1 + f_{n-1})^{\frac{1}{n-1}}
   \]

   Combining these inequalities:
   \[
   (1 + f_n)^{\frac{1}{n}} < (1 + f_{n-1})^{\frac{1}{n-1}}
   \]

   Therefore:
   \[
   3 \sqrt[n]{1 + f_n} < 3 \sqrt[n-1]{1 + f_{n-1}}
   \]

   Hence:
   \[
   a_n < a_{n-1} \quad \forall n \geq 2
   \]

   This proves that \( \{a_n\} \) is a decreasing sequence.

   \(\blacksquare\)