The number of integral terms in the expansion of `(sqrt3+ root8 5)^(256` is

Options

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

Answer

To find the number of integral terms in the expansion of \((\sqrt{3} + \sqrt[8]{5})^{256}\), we can follow these steps:

### Step 1: Identify the General Term
The general term \(T_r\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = \sqrt{3}\), \(b = \sqrt[8]{5}\), and \(n = 256\). Therefore, the general term becomes:
\[ T_r = \binom{256}{r} (\sqrt{3})^{256 - r} \left(\sqrt[8]{5}\right)^r \]

### Step 2: Simplify the General Term
We can simplify \(T_r\):
\[ T_r = \binom{256}{r} (3^{\frac{256 - r}{2}}) (5^{\frac{r}{8}}) \]
This means:
\[ T_r = \binom{256}{r} \cdot 3^{\frac{256 - r}{2}} \cdot 5^{\frac{r}{8}} \]

### Step 3: Determine Conditions for Integral Terms
For \(T_r\) to be an integer, both exponents \(\frac{256 - r}{2}\) and \(\frac{r}{8}\) must be integers. This leads us to the following conditions:
1. \(\frac{256 - r}{2}\) is an integer, which implies \(256 - r\) must be even. Therefore, \(r\) must also be even.
2. \(\frac{r}{8}\) is an integer, which implies \(r\) must be a multiple of 8.

### Step 4: Find Possible Values of \(r\)
Let \(r = 8k\) where \(k\) is an integer. Since \(r\) must be even, we need to find the values of \(k\) such that:
\[ 0 \leq r \leq 256 \implies 0 \leq 8k \leq 256 \implies 0 \leq k \leq 32 \]
Thus, \(k\) can take values from \(0\) to \(32\).

### Step 5: Count the Integral Terms
The possible values of \(k\) are \(0, 1, 2, \ldots, 32\). This gives us:
\[ k = 0, 1, 2, \ldots, 32 \implies \text{Total values} = 32 - 0 + 1 = 33 \]

### Conclusion
The number of integral terms in the expansion of \((\sqrt{3} + \sqrt[8]{5})^{256}\) is **33**.

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