If `(1+x+2x^2)^20 = a_0+a_1x+a_2x^2+. . . +a_40x^40` then `a_1+a_3+a_5+ . . . +a_37=`

Answer

To solve the problem, we need to find the sum of the coefficients of the odd powers of \( x \) in the expansion of \( (1 + x + 2x^2)^{20} \). Let's denote the expansion as:

\[ (1 + x + 2x^2)^{20} = a_0 + a_1 x + a_2 x^2 + \ldots + a_{40} x^{40} \]

We are interested in finding \( a_1 + a_3 + a_5 + \ldots + a_{37} \).

### Step 1: Calculate \( (1 + x + 2x^2)^{20} \) at \( x = 1 \)

First, we evaluate the expression at \( x = 1 \):

\[ (1 + 1 + 2 \cdot 1^2)^{20} = (1 + 1 + 2)^{20} = 4^{20} \]

This gives us the sum of all coefficients:

\[ a_0 + a_1 + a_2 + \ldots + a_{40} = 4^{20} \]

### Step 2: Calculate \( (1 + x + 2x^2)^{20} \) at \( x = -1 \)

Next, we evaluate the expression at \( x = -1 \):

\[ (1 - 1 + 2(-1)^2)^{20} = (1 - 1 + 2)^{20} = 2^{20} \]

This gives us the sum of the coefficients with alternating signs:

\[ a_0 - a_1 + a_2 - a_3 + \ldots + a_{40} = 2^{20} \]

### Step 3: Set up the equations

We now have two equations:

1. \( a_0 + a_1 + a_2 + \ldots + a_{40} = 4^{20} \)
2. \( a_0 - a_1 + a_2 - a_3 + \ldots + a_{40} = 2^{20} \)

### Step 4: Add the two equations

Adding these two equations will help us isolate the odd coefficients:

\[ (a_0 + a_1 + a_2 + \ldots + a_{40}) + (a_0 - a_1 + a_2 - a_3 + \ldots + a_{40}) = 4^{20} + 2^{20} \]

This simplifies to:

\[ 2a_0 + 2a_2 + 2a_4 + \ldots + 2a_{40} = 4^{20} + 2^{20} \]

### Step 5: Subtract the second equation from the first

Now, subtract the second equation from the first:

\[ (a_0 + a_1 + a_2 + \ldots + a_{40}) - (a_0 - a_1 + a_2 - a_3 + \ldots + a_{40}) = 4^{20} - 2^{20} \]

This simplifies to:

\[ 2(a_1 + a_3 + a_5 + \ldots + a_{39}) = 4^{20} - 2^{20} \]

### Step 6: Solve for the sum of odd coefficients

Now we can isolate the sum of the odd coefficients:

\[ a_1 + a_3 + a_5 + \ldots + a_{39} = \frac{4^{20} - 2^{20}}{2} \]

### Step 7: Calculate the final result

Now we compute \( 4^{20} \) and \( 2^{20} \):

\[ 4^{20} = (2^2)^{20} = 2^{40} \]
\[ 2^{20} = 2^{20} \]

Thus, we have:

\[ a_1 + a_3 + a_5 + \ldots + a_{39} = \frac{2^{40} - 2^{20}}{2} = 2^{39} - 2^{19} \]

### Final Answer

The value of \( a_1 + a_3 + a_5 + \ldots + a_{37} \) is:

\[ \boxed{2^{39} - 2^{19}} \]