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Solution

Key Concept: AM-GM Inequality
$\dfrac{a_1 + a_2 + \cdots + a_n}{n} \geq (a_1 \cdot a_2 \cdots a_n)^{1/n}$

Step 1: Rewrite the sum
$a_1 + a_2 + \cdots + a_{n-1} + 2a_n$

This has $n$ terms: $(a_1, a_2, \ldots, a_{n-1}, 2a_n)$

Step 2: Apply AM-GM
$\dfrac{a_1 + a_2 + \cdots + a_{n-1} + 2a_n}{n} \geq (a_1 \cdot a_2 \cdots a_{n-1} \cdot 2a_n)^{1/n}$

$\geq (2 \cdot a_1 a_2 \cdots a_n)^{1/n}$

$\geq (2c)^{1/n}$

Step 3: Find minimum
$a_1 + a_2 + \cdots + 2a_n \geq n(2c)^{1/n}$

Minimum value $= n(2c)^{1/n}$

Equality holds when $a_1 = a_2 = \cdots = a_{n-1} = 2a_n$

Answer: Minimum value $= \boxed{n(2c)^{1/n}}$