The numberof integral termsin the expansion of `( (3)-root(8)(5))^256` is (A) 32 (B) 33 (C) 34 (D) 35

Options

A. 32
B. 33
C. 34
D. 35

Answer

To find the number of integral terms in the expansion of \( \left( \sqrt{3} - 8\sqrt{5} \right)^{256} \), we will use the Binomial Theorem. Let's break down the solution step by step.

### Step 1: Rewrite the expression
The expression can be rewritten as:
\[ \left( 3^{1/2} - 8 \cdot 5^{1/2} \right)^{256} \]

### Step 2: Apply the Binomial Theorem
According to the Binomial Theorem, the expansion of \( (a + b)^n \) is given by:
\[ \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = \sqrt{3} \) and \( b = -8\sqrt{5} \), and \( n = 256 \). Thus, the expansion becomes:
\[ \sum_{r=0}^{256} \binom{256}{r} (\sqrt{3})^{256-r} (-8\sqrt{5})^r \]

### Step 3: Simplify the terms
Each term in the expansion can be expressed as:
\[ \binom{256}{r} 3^{(256-r)/2} (-8)^r 5^{r/2} \]
This simplifies to:
\[ \binom{256}{r} (-8)^r 3^{(256-r)/2} 5^{r/2} \]

### Step 4: Determine conditions for integral terms
For the term to be integral, both exponents \( \frac{256 - r}{2} \) and \( \frac{r}{2} \) must be integers. This implies:
1. \( 256 - r \) must be even, which means \( r \) must be even.
2. \( r \) must also be even.

Let \( r = 2k \) where \( k \) is an integer. Then, substituting \( r \) gives:
\[ 256 - 2k \text{ must be even} \]
This condition is satisfied since \( 256 \) is even.

### Step 5: Find the range of \( r \)
Since \( r \) can take values from \( 0 \) to \( 256 \) and must be even, the possible values for \( r \) are:
\[ 0, 2, 4, \ldots, 256 \]
This is an arithmetic series where:
- First term \( a = 0 \)
- Last term \( l = 256 \)
- Common difference \( d = 2 \)

### Step 6: Calculate the number of terms
The number of terms \( n \) in this series can be calculated using the formula for the \( n \)-th term of an arithmetic series:
\[ l = a + (n-1)d \]
Substituting the known values:
\[ 256 = 0 + (n-1) \cdot 2 \]
Solving for \( n \):
\[ 256 = 2(n-1) \implies 128 = n - 1 \implies n = 129 \]

### Step 7: Count the integral terms
Since \( r \) must be even, the possible values of \( r \) are \( 0, 2, 4, \ldots, 256 \), which gives us \( 129 \) possible values.

### Final Count
Thus, the number of integral terms in the expansion of \( \left( \sqrt{3} - 8\sqrt{5} \right)^{256} \) is:
\[ \text{Number of integral terms} = 129 \]

### Conclusion
The correct answer is not among the options provided (32, 33, 34, 35). The number of integral terms is 129.