The number of integral terms in the expansion of `(sqrt2 + root(4)(3))^100` is

Options

A. 75
B. 25
C. 26
D. 101

Answer

To find the number of integral terms in the expansion of \((\sqrt{2} + \sqrt[4]{3})^{100}\), we will follow these steps:

### Step 1: Rewrite the expression
The expression can be rewritten as:
\[ (\sqrt{2} + \sqrt[4]{3})^{100} = (2^{1/2} + 3^{1/4})^{100} \]

### Step 2: Identify the general term
Using the Binomial Theorem, the general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = \sqrt{2}\), \(b = \sqrt[4]{3}\), and \(n = 100\). Thus, the general term becomes:
\[ T_{r+1} = \binom{100}{r} (\sqrt{2})^{100-r} (\sqrt[4]{3})^r \]
This simplifies to:
\[ T_{r+1} = \binom{100}{r} (2^{(100-r)/2}) (3^{r/4}) \]

### Step 3: Determine when the term is integral
For \(T_{r+1}\) to be an integer, both \(2^{(100-r)/2}\) and \(3^{r/4}\) must be integers.

1. **Condition for \(2^{(100-r)/2}\)**:
- \((100 - r)\) must be even, which implies \(r\) must be even.

2. **Condition for \(3^{r/4}\)**:
- \(r\) must be a multiple of 4.

### Step 4: Find the values of \(r\)
Since \(r\) must be both even and a multiple of 4, we can express \(r\) as:
\[ r = 4k \quad \text{for integers } k \]
Now, since \(r\) can take values from 0 to 100, we need to find the possible values of \(k\):
\[ 0 \leq 4k \leq 100 \implies 0 \leq k \leq 25 \]
Thus, \(k\) can take values \(0, 1, 2, \ldots, 25\).

### Step 5: Count the integral terms
The possible values of \(k\) are \(0\) through \(25\), which gives us a total of:
\[ 25 - 0 + 1 = 26 \text{ integral terms} \]

### Final Answer
The number of integral terms in the expansion of \((\sqrt{2} + \sqrt[4]{3})^{100}\) is \(26\).

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