Number of integral terms in the expansion of `(sqrt5+root8(7))^(1024)` is

Answer

To find the number of integral terms in the expansion of \((\sqrt{5} + \sqrt[8]{7})^{1024}\), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \(T_r\) in the expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = \sqrt{5}\), \(b = \sqrt[8]{7}\), and \(n = 1024\). Therefore, the general term becomes:
\[ T_r = \binom{1024}{r} (\sqrt{5})^{1024-r} (\sqrt[8]{7})^r \]

### Step 2: Simplify the general term
We can rewrite the general term as follows:
\[ T_r = \binom{1024}{r} (5^{\frac{1024-r}{2}}) (7^{\frac{r}{8}}) \]
This means:
\[ T_r = \binom{1024}{r} \cdot 5^{\frac{1024-r}{2}} \cdot 7^{\frac{r}{8}} \]

### Step 3: Determine conditions for \(T_r\) to be an integer
For \(T_r\) to be an integer, both \(5^{\frac{1024-r}{2}}\) and \(7^{\frac{r}{8}}\) must be integers. This leads us to two conditions:
1. \(\frac{1024 - r}{2}\) must be an integer, which implies \(1024 - r\) must be even. Therefore, \(r\) must be even.
2. \(\frac{r}{8}\) must be an integer, which implies \(r\) must be a multiple of \(8\).

### Step 4: Find the range of \(r\)
Since \(r\) must be even and a multiple of \(8\), we can express \(r\) as:
\[ r = 8k \quad \text{for some integer } k \]
Now, \(r\) can take values from \(0\) to \(1024\). Thus, we need to find the possible values of \(k\):
\[ 0 \leq 8k \leq 1024 \implies 0 \leq k \leq \frac{1024}{8} = 128 \]
This means \(k\) can take values from \(0\) to \(128\).

### Step 5: Count the number of integral terms
The possible values of \(k\) are \(0, 1, 2, \ldots, 128\). This gives us a total of:
\[ 128 - 0 + 1 = 129 \]
Thus, the number of integral terms in the expansion is \(129\).

### Final Answer
The number of integral terms in the expansion of \((\sqrt{5} + \sqrt[8]{7})^{1024}\) is \(129\).

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