In the expansion of `(3sqrt(a/b)+3sqrt(b/sqrt(a)))^21`, the term containing same powers of a & b is

Options

A. `11^("th")`
B. `13^("th")`
C. `12^("th")`
D. `6^("th)`

Answer

To find the term containing the same powers of \( a \) and \( b \) in the expansion of \( (3\sqrt{\frac{a}{b}} + 3\sqrt{\frac{b}{\sqrt{a}}})^{21} \), we can follow these steps:

### Step 1: Write the General Term
The general term \( T_{r+1} \) in the binomial expansion is given by:
\[ T_{r+1} = \binom{21}{r} \left(3\sqrt{\frac{a}{b}}\right)^{21-r} \left(3\sqrt{\frac{b}{\sqrt{a}}}\right)^{r} \]

### Step 2: Simplify the General Term
Now, we simplify the general term:
\[ T_{r+1} = \binom{21}{r} \cdot 3^{21} \cdot \left(\frac{a^{1/2}}{b^{1/2}}\right)^{21-r} \cdot \left(\frac{b^{1/2}}{a^{1/4}}\right)^{r} \]
This simplifies to:
\[ T_{r+1} = \binom{21}{r} \cdot 3^{21} \cdot a^{\frac{21-r}{2}} \cdot b^{-\frac{21-r}{2}} \cdot b^{\frac{r}{2}} \cdot a^{-\frac{r}{4}} \]
Combining the powers of \( a \) and \( b \):
\[ T_{r+1} = \binom{21}{r} \cdot 3^{21} \cdot a^{\frac{21-r}{2} - \frac{r}{4}} \cdot b^{-\frac{21-r}{2} + \frac{r}{2}} \]

### Step 3: Combine the Exponents
The exponent of \( a \) is:
\[ \frac{21 - r}{2} - \frac{r}{4} = \frac{42 - 2r - r}{4} = \frac{42 - 3r}{4} \]
The exponent of \( b \) is:
\[ -\frac{21 - r}{2} + \frac{r}{2} = \frac{-21 + r + r}{2} = \frac{-21 + 2r}{2} \]

### Step 4: Set the Exponents Equal
To find the term where the powers of \( a \) and \( b \) are the same, we set the exponents equal:
\[ \frac{42 - 3r}{4} = \frac{-21 + 2r}{2} \]

### Step 5: Solve for \( r \)
Cross-multiplying gives:
\[ 42 - 3r = -42 + 4r \]
Combining like terms:
\[ 42 + 42 = 4r + 3r \implies 84 = 7r \implies r = 12 \]

### Step 6: Find the Term
The term containing the same powers of \( a \) and \( b \) is the \( (r+1)^{th} \) term, which is:
\[ T_{r+1} = T_{13} \]

### Conclusion
Thus, the term containing the same powers of \( a \) and \( b \) is the 13th term in the expansion.