The first integral term in the expansion of `(sqrt3+2^(1/3))^9`, is

Options

A. `2^(nd) `term
B. `3^(rd)` term
C. `4^(th)` term
D. `5^(th)` term

Answer

To find the first integral term in the expansion of \((\sqrt{3} + 2^{1/3})^9\), we can use the Binomial Theorem, which states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

Here, \(a = \sqrt{3}\), \(b = 2^{1/3}\), and \(n = 9\).

### Step 1: Write the general term
The general term \(T_k\) in the expansion is given by:

\[ T_k = \binom{9}{k} (\sqrt{3})^{9-k} (2^{1/3})^k \]

### Step 2: Simplify the general term
We can simplify \(T_k\):

\[ T_k = \binom{9}{k} (3^{1/2})^{9-k} (2^{1/3})^k = \binom{9}{k} 3^{(9-k)/2} 2^{k/3} \]

### Step 3: Find conditions for integral terms
For \(T_k\) to be an integral term, both exponents \((9-k)/2\) and \(k/3\) must be integers.

1. **Condition for \(3^{(9-k)/2}\)**:
\((9-k)/2\) is an integer if \(9-k\) is even, which implies \(k\) must be odd.

2. **Condition for \(2^{k/3}\)**:
\(k/3\) is an integer if \(k\) is a multiple of 3.

### Step 4: Find suitable values of \(k\)
We need \(k\) to be both odd and a multiple of 3. The smallest odd multiple of 3 is \(3\).

### Step 5: Calculate the fourth term
Now, we can calculate \(T_3\):

\[ T_3 = \binom{9}{3} 3^{(9-3)/2} 2^{3/3} \]

Calculating each part:

- \(\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\)
- \(3^{(9-3)/2} = 3^{6/2} = 3^3 = 27\)
- \(2^{3/3} = 2^1 = 2\)

Now substituting these values into \(T_3\):

\[ T_3 = 84 \times 27 \times 2 \]

### Step 6: Calculate the final result
Calculating \(T_3\):

\[ T_3 = 84 \times 27 = 2268 \]
\[ T_3 = 2268 \times 2 = 4536 \]

Thus, the first integral term in the expansion of \((\sqrt{3} + 2^{1/3})^9\) is **4536**.