Find the sum of the coefficients of integral powers
of x in `(1+3sqrtx)^20`

Answer

To find the sum of the coefficients of integral powers of \( x \) in the expression \( (1 + 3\sqrt{x})^{20} \), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = 1 \), \( b = 3\sqrt{x} \), and \( n = 20 \). Therefore, the general term becomes:
\[ T_{r+1} = \binom{20}{r} \cdot 1^{20-r} \cdot (3\sqrt{x})^r = \binom{20}{r} \cdot 3^r \cdot (\sqrt{x})^r \]

### Step 2: Simplify the term
Since \( \sqrt{x} = x^{1/2} \), we can rewrite the term as:
\[ T_{r+1} = \binom{20}{r} \cdot 3^r \cdot x^{r/2} \]

### Step 3: Determine conditions for integral powers of \( x \)
For \( x^{r/2} \) to be an integral power of \( x \), \( r/2 \) must be an integer. This implies that \( r \) must be an even integer. Therefore, we can let \( r = 2k \) where \( k \) is a non-negative integer.

### Step 4: Substitute \( r \) in the general term
Substituting \( r = 2k \) into the general term gives:
\[ T_{2k+1} = \binom{20}{2k} \cdot 3^{2k} \cdot x^{k} \]

### Step 5: Find the sum of coefficients for integral powers
We need to sum the coefficients of \( x^k \) for \( k = 0, 1, 2, \ldots, 10 \) (since \( r \) can go up to 20, \( k \) can go up to 10):
\[ \text{Sum of coefficients} = \sum_{k=0}^{10} \binom{20}{2k} \cdot 3^{2k} \]

### Step 6: Use the binomial theorem to evaluate the sum
To evaluate this sum, we can use the identity:
\[ (1 + x)^n + (1 - x)^n = 2 \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} x^{2k} \]
Setting \( n = 20 \) and \( x = 3 \):
\[ (1 + 3)^{20} + (1 - 3)^{20} = 4^{20} + (-2)^{20} \]
Calculating:
\[ 4^{20} = (2^2)^{20} = 2^{40}, \quad (-2)^{20} = 2^{20} \]
Thus:
\[ 2^{40} + 2^{20} = 2^{20}(2^{20} + 1) \]

### Step 7: Divide by 2 to find the sum of coefficients
Now, we divide by 2 to find the sum of coefficients:
\[ \text{Sum of coefficients} = \frac{1}{2} \left( 2^{40} + 2^{20} \right) = 2^{39} + 2^{19} \]

### Final Result
Therefore, the sum of the coefficients of integral powers of \( x \) in \( (1 + 3\sqrt{x})^{20} \) is:
\[ \boxed{2^{39} + 2^{19}} \]