The two consecutive terms whose coefficients in the expansion of `(3 + 2x)^(94)` are equal, are

Options

A. 37th, 38th
B. 38th, 39th
C. 47th, 48th
D. 48th, 49th

Answer

To find the two consecutive terms in the expansion of \((3 + 2x)^{94}\) whose coefficients are equal, we can follow these steps:

### Step 1: Write the General Term
The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our case, \(a = 3\), \(b = 2x\), and \(n = 94\). Therefore, the general term becomes:
\[ T_{r+1} = \binom{94}{r} (3)^{94-r} (2x)^r = \binom{94}{r} 3^{94-r} 2^r x^r \]

### Step 2: Write the Coefficients of Two Consecutive Terms
The coefficients of two consecutive terms \(T_{r+1}\) and \(T_{r+2}\) are:
- Coefficient of \(T_{r+1}\): \(\binom{94}{r} 3^{94-r} 2^r\)
- Coefficient of \(T_{r+2}\): \(\binom{94}{r+1} 3^{94-(r+1)} 2^{r+1}\)

### Step 3: Set the Coefficients Equal
We need to set the coefficients of these two terms equal:
\[ \binom{94}{r} 3^{94-r} 2^r = \binom{94}{r+1} 3^{93-r} 2^{r+1} \]

### Step 4: Simplify the Equation
Using the property of binomial coefficients:
\[ \binom{94}{r+1} = \frac{94 - r}{r + 1} \binom{94}{r} \]
Substituting this into our equation gives:
\[ \binom{94}{r} 3^{94-r} 2^r = \frac{94 - r}{r + 1} \binom{94}{r} 3^{93-r} 2^{r+1} \]
We can cancel \(\binom{94}{r}\) from both sides (assuming \(r \neq 0\)):
\[ 3^{94-r} 2^r = \frac{94 - r}{r + 1} 3^{93-r} 2^{r+1} \]

### Step 5: Further Simplification
Rearranging gives:
\[ 3 \cdot 3^{93-r} 2^r = \frac{94 - r}{r + 1} 3^{93-r} 2^{r+1} \]
Dividing both sides by \(3^{93-r}\) and \(2^r\) (assuming \(r \neq 0\)):
\[ 3 = \frac{94 - r}{r + 1} \cdot 2 \]

### Step 6: Cross Multiply
Cross multiplying gives:
\[ 3(r + 1) = 2(94 - r) \]
Expanding both sides:
\[ 3r + 3 = 188 - 2r \]

### Step 7: Solve for \(r\)
Rearranging gives:
\[ 3r + 2r = 188 - 3 \]
\[ 5r = 185 \]
\[ r = 37 \]

### Step 8: Find Consecutive Terms
The two consecutive terms are \(T_{r+1}\) and \(T_{r+2}\):
- For \(r = 37\), the terms are:
- \(T_{38} = \binom{94}{37} 3^{57} (2x)^{37}\)
- \(T_{39} = \binom{94}{38} 3^{56} (2x)^{38}\)

### Conclusion
The two consecutive terms whose coefficients are equal in the expansion of \((3 + 2x)^{94}\) are \(T_{38}\) and \(T_{39}\).

---