If the (r + 1)th term in the expansion of `((a^(1//3))/(b^(1//6))+(b^(1//2))/(a^(1//6)))^(21)` has equal exponents of both a and b, then value of r is

Options

A. 8
B. 9
C. 10
D. 11

Answer

To solve the problem, we need to find the value of \( r \) such that the \( (r + 1) \)th term in the expansion of

\[ \left( \frac{a^{1/3}}{b^{1/6}} + \frac{b^{1/2}}{a^{1/6}} \right)^{21} \]

has equal exponents of \( a \) and \( b \).

### Step-by-step Solution:

1. **Identify the general term in the binomial expansion**:
The general term \( T_{r+1} \) in the expansion of \( (x + y)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]
Here, \( n = 21 \), \( x = \frac{a^{1/3}}{b^{1/6}} \), and \( y = \frac{b^{1/2}}{a^{1/6}} \).

2. **Substituting values into the general term**:
Thus, we have:
\[ T_{r+1} = \binom{21}{r} \left( \frac{a^{1/3}}{b^{1/6}} \right)^{21-r} \left( \frac{b^{1/2}}{a^{1/6}} \right)^{r} \]

3. **Simplifying the term**:
Expanding this, we get:
\[ T_{r+1} = \binom{21}{r} a^{(21-r)/3} b^{-(21-r)/6} b^{r/2} a^{-r/6} \]
Combining the powers of \( a \) and \( b \):
\[ T_{r+1} = \binom{21}{r} a^{\frac{21-r}{3} - \frac{r}{6}} b^{\frac{r}{2} - \frac{21-r}{6}} \]

4. **Finding the exponents**:
The exponent of \( a \) is:
\[ \frac{21-r}{3} - \frac{r}{6} = \frac{42 - 2r - r}{6} = \frac{42 - 3r}{6} \]
The exponent of \( b \) is:
\[ \frac{r}{2} - \frac{21-r}{6} = \frac{3r - (21 - r)}{6} = \frac{4r - 21}{6} \]

5. **Setting the exponents equal**:
We need to set the exponents of \( a \) and \( b \) equal:
\[ \frac{42 - 3r}{6} = \frac{4r - 21}{6} \]
Multiplying through by 6 to eliminate the denominator:
\[ 42 - 3r = 4r - 21 \]

6. **Solving for \( r \)**:
Rearranging gives:
\[ 42 + 21 = 4r + 3r \]
\[ 63 = 7r \]
\[ r = 9 \]

### Final Answer:
Thus, the value of \( r \) is \( 9 \).