Total number of integral terms in `(4^(1/4)+5^(1/6))^(120)` is ________

Answer

To find the total number of integral terms in the expression \((4^{1/4} + 5^{1/6})^{120}\), we will use the Binomial Theorem. Let's break down the solution step by step.

### Step 1: Apply the Binomial Theorem
According to the Binomial Theorem, we can expand \((a + b)^n\) as follows:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
In our case, let \(a = 4^{1/4}\) and \(b = 5^{1/6}\), and \(n = 120\). Thus, we have:
\[ (4^{1/4} + 5^{1/6})^{120} = \sum_{r=0}^{120} \binom{120}{r} (4^{1/4})^{120 - r} (5^{1/6})^r \]

### Step 2: Simplify the Terms
Now, we simplify the terms:
\[ (4^{1/4})^{120 - r} = 4^{(120 - r)/4} = 2^{2(120 - r)/4} = 2^{(240 - 2r)/4} = 2^{60 - \frac{r}{2}} \]
\[ (5^{1/6})^r = 5^{r/6} \]
Thus, the general term becomes:
\[ \binom{120}{r} 2^{60 - \frac{r}{2}} 5^{\frac{r}{6}} \]

### Step 3: Determine Conditions for Integral Terms
For the term to be an integer, both \(2^{60 - \frac{r}{2}}\) and \(5^{\frac{r}{6}}\) must be integers. This leads to the following conditions:
1. \(60 - \frac{r}{2} \geq 0\) which implies \(r \leq 120\).
2. \(\frac{r}{6}\) must be an integer, which implies \(r\) must be a multiple of 6.

Let \(r = 6k\) where \(k\) is a non-negative integer.

### Step 4: Substitute and Solve for \(k\)
Substituting \(r = 6k\) into the inequality:
\[ 60 - \frac{6k}{2} \geq 0 \implies 60 - 3k \geq 0 \implies 3k \leq 60 \implies k \leq 20 \]
Thus, \(k\) can take values from \(0\) to \(20\).

### Step 5: Count the Integral Terms
The possible values of \(k\) are \(0, 1, 2, \ldots, 20\), which gives us a total of:
\[ 20 - 0 + 1 = 21 \]
Thus, the total number of integral terms in the expansion of \((4^{1/4} + 5^{1/6})^{120}\) is **21**.

### Final Answer
The total number of integral terms is **21**.