The sum of all rational terms in the expansion of
` (3^(1//4) + 4^(1//3))^(12)` is

Options

A. 91
B. 251
C. 273
D. 283

Explanation

To find the sum of all rational terms in the expansion of \( (3^{1/4} + 4^{1/3})^{12} \), we will follow these steps:

### Step 1: Write the General Term
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, \( a = 3^{1/4} \), \( b = 4^{1/3} \), and \( n = 12 \). Therefore, the general term becomes:

\[ T_{r+1} = \binom{12}{r} (3^{1/4})^{12-r} (4^{1/3})^r \]

### Step 2: Simplify the General Term
Now, simplifying \( T_{r+1} \):

\[ T_{r+1} = \binom{12}{r} \cdot 3^{(12-r)/4} \cdot 4^{r/3} \]

### Step 3: Determine Conditions for Rational Terms
For \( T_{r+1} \) to be a rational term, both exponents of 3 and 4 must be integers. This requires:

1. \( \frac{12 - r}{4} \) to be an integer, which implies \( 12 - r \) must be divisible by 4.
2. \( \frac{r}{3} \) to be an integer, which implies \( r \) must be divisible by 3.

### Step 4: Find Values of \( r \)
Let’s analyze the conditions:

- From \( 12 - r \equiv 0 \mod 4 \), we have possible values for \( r \) as \( 0, 4, 8, 12 \).
- From \( r \equiv 0 \mod 3 \), we have possible values for \( r \) as \( 0, 3, 6, 9, 12 \).

The common values satisfying both conditions are \( r = 0 \) and \( r = 12 \).

### Step 5: Calculate Rational Terms
Now we will calculate the rational terms for \( r = 0 \) and \( r = 12 \):

1. For \( r = 0 \):
\[ T_1 = \binom{12}{0} (3^{1/4})^{12} (4^{1/3})^0 = 1 \cdot 3^{3} \cdot 1 = 27 \]

2. For \( r = 12 \):
\[ T_{13} = \binom{12}{12} (3^{1/4})^{0} (4^{1/3})^{12} = 1 \cdot 1 \cdot 4^{4} = 256 \]

### Step 6: Sum of Rational Terms
Now, we sum the rational terms:

\[ \text{Sum} = T_1 + T_{13} = 27 + 256 = 283 \]

### Final Answer
Thus, the sum of all rational terms in the expansion is:

\[ \boxed{283} \]