Term independent of x in `(1+4x)^p (1+(1)/(4x) )^q` is :

Options

A. `""^((p+q))C_(4p)`
B. `""^((p+q))C_p`
C. `""^((p+q))C_(2p)`
D. `""^((p+q))C_(3p)`

Answer

To find the term independent of \( x \) in the expression \( (1 + 4x)^p \cdot \left(1 + \frac{1}{4x}\right)^q \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (1 + 4x)^p \cdot \left(1 + \frac{1}{4x}\right)^q \]

### Step 2: Identify the general term
The general term of \( (1 + 4x)^p \) can be expressed as:
\[ T_r = \binom{p}{r} (4x)^r = \binom{p}{r} 4^r x^r \]
And the general term of \( \left(1 + \frac{1}{4x}\right)^q \) can be expressed as:
\[ T_s = \binom{q}{s} \left(\frac{1}{4x}\right)^s = \binom{q}{s} \frac{1}{(4x)^s} = \binom{q}{s} \frac{1}{4^s} x^{-s} \]

### Step 3: Combine the general terms
The combined general term for the product is:
\[ T_{r,s} = \binom{p}{r} 4^r x^r \cdot \binom{q}{s} \frac{1}{4^s} x^{-s} = \binom{p}{r} \binom{q}{s} \frac{4^r}{4^s} x^{r-s} \]

### Step 4: Find the condition for the term to be independent of \( x \)
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ r - s = 0 \quad \Rightarrow \quad r = s \]

### Step 5: Substitute \( s = r \) into the general term
Substituting \( s = r \) into the general term gives:
\[ T_{r,r} = \binom{p}{r} \binom{q}{r} \frac{4^r}{4^r} = \binom{p}{r} \binom{q}{r} \]

### Step 6: Determine the range of \( r \)
The value of \( r \) can range from \( 0 \) to \( \min(p, q) \).

### Step 7: Final expression for the term independent of \( x \)
Thus, the term independent of \( x \) is given by:
\[ \sum_{r=0}^{\min(p,q)} \binom{p}{r} \binom{q}{r} \]

### Conclusion
The term independent of \( x \) in the expansion of \( (1 + 4x)^p \cdot \left(1 + \frac{1}{4x}\right)^q \) is:
\[ \sum_{r=0}^{\min(p,q)} \binom{p}{r} \binom{q}{r} \]