To solve the given problem, we will analyze both statements step by step.
### Step 1: Analyze Statement 1
We are given the equation:
\[ a + 2b + 3c = 1 \]
with the conditions \( a > 0 \), \( b > 0 \), and \( c > 0 \). We need to find the maximum value of the expression:
\[ E = a^3 b^2 c \]
### Step 2: Apply the Method of Lagrange Multipliers
To maximize \( E \) subject to the constraint \( g(a, b, c) = a + 2b + 3c - 1 = 0 \), we can use the method of Lagrange multipliers. We set up the equations:
\[
\nabla E = \lambda \nabla g
\]
where \( \nabla E = \left( \frac{\partial E}{\partial a}, \frac{\partial E}{\partial b}, \frac{\partial E}{\partial c} \right) \) and \( \nabla g = \left( \frac{\partial g}{\partial a}, \frac{\partial g}{\partial b}, \frac{\partial g}{\partial c} \right) \).
Calculating the gradients:
\[
\frac{\partial E}{\partial a} = 3a^2 b^2 c, \quad \frac{\partial E}{\partial b} = 2a^3 b c, \quad \frac{\partial E}{\partial c} = a^3 b^2
\]
\[
\frac{\partial g}{\partial a} = 1, \quad \frac{\partial g}{\partial b} = 2, \quad \frac{\partial g}{\partial c} = 3
\]
Setting up the equations:
\[
3a^2 b^2 c = \lambda, \quad 2a^3 b c = 2\lambda, \quad a^3 b^2 = 3\lambda
\]
### Step 3: Solve the Equations
From the equations, we can express \( \lambda \):
1. From \( 3a^2 b^2 c = \lambda \)
2. From \( 2a^3 b c = 2\lambda \) implies \( \lambda = a^3 b c \)
3. From \( a^3 b^2 = 3\lambda \)
Setting the expressions for \( \lambda \) equal to each other:
\[
3a^2 b^2 c = a^3 b c \implies 3b = a \quad (1)
\]
\[
3a^2 b^2 c = \frac{1}{3} a^3 b^2 \implies 3c = a \quad (2)
\]
### Step 4: Substitute Back into the Constraint
Substituting \( a = 3b \) and \( a = 3c \) into the constraint:
\[
3b + 2b + 3\left(\frac{b}{3}\right) = 1 \implies 3b + 2b + b = 1 \implies 6b = 1 \implies b = \frac{1}{6}
\]
Then,
\[
a = 3b = \frac{1}{2}, \quad c = \frac{b}{3} = \frac{1}{18}
\]
### Step 5: Calculate Maximum Value of \( E \)
Now substituting \( a, b, c \) back into \( E \):
\[
E = \left(\frac{1}{2}\right)^3 \left(\frac{1}{6}\right)^2 \left(\frac{1}{18}\right) = \frac{1}{8} \cdot \frac{1}{36} \cdot \frac{1}{18} = \frac{1}{5184}
\]
Thus, the maximum value of \( a^3 b^2 c \) is indeed \( \frac{1}{5184} \).
### Conclusion for Statement 1
Statement 1 is true.
### Step 6: Analyze Statement 2
The second statement claims that there exists an arithmetic progression such that the sum of its \( n \) terms is given by:
\[ S_n = an^3 + bn^2 + cn + d \]
### Step 7: Determine Validity of Statement 2
The sum of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} (2a + (n-1)d) \]
This is a linear function of \( n \), not a cubic polynomial. Therefore, Statement 2 is false.
### Final Answer
- Statement 1 is true.
- Statement 2 is false.
