If the number of terms free from radicals in the expansion of `(7^((1)/(3))+11^((1)/(9)))^(6561)` is k, then the value of `(k)/(100)` is equal to

Answer

To solve the problem, we need to find the number of terms free from radicals in the expansion of \((7^{\frac{1}{3}} + 11^{\frac{1}{9}})^{6561}\).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our expression, \(a = 7^{\frac{1}{3}}\), \(b = 11^{\frac{1}{9}}\), and \(n = 6561\). Thus, the general term becomes:
\[ T_{r+1} = \binom{6561}{r} (7^{\frac{1}{3}})^{6561 - r} (11^{\frac{1}{9}})^{r} \]
Simplifying this gives:
\[ T_{r+1} = \binom{6561}{r} 7^{\frac{6561 - r}{3}} 11^{\frac{r}{9}} \]

2. **Condition for Rational Terms**:
We want the term to be free from radicals, which means both exponents must be integers. Therefore, we need:
- \(\frac{6561 - r}{3}\) to be an integer, which implies \(6561 - r \equiv 0 \mod 3\)
- \(\frac{r}{9}\) to be an integer, which implies \(r \equiv 0 \mod 9\)

3. **Finding Valid Values for \(r\)**:
From \(r \equiv 0 \mod 9\), we can write \(r = 9k\) for some integer \(k\). The maximum value of \(r\) is \(6561\), so:
\[ 9k \leq 6561 \implies k \leq \frac{6561}{9} = 729 \]
Therefore, \(k\) can take values from \(0\) to \(729\).

4. **Counting Valid \(k\) Values**:
The possible values of \(k\) are \(0, 1, 2, \ldots, 729\). This gives us a total of:
\[ 729 - 0 + 1 = 730 \text{ values} \]

5. **Finding \(k/100\)**:
Since the number of terms free from radicals is \(k = 730\), we need to find:
\[ \frac{k}{100} = \frac{730}{100} = 7.3 \]

### Final Answer:
Thus, the value of \(\frac{k}{100}\) is \(7.3\).