Numerically the longest term in the expansion of `(3+2x)^(50)`, when `x=(1)/( 5)` is

Options

A. `6^(th)`
B. `8^(th)`
C. `7^(th)`
D. None of these

Answer

To find the numerically longest term in the expansion of \((3 + 2x)^{50}\) when \(x = \frac{1}{5}\), we will follow these steps:

### Step 1: Identify the General Term
The general term \(T_k\) in the expansion of \((a + b)^n\) is given by:
\[ T_k = \binom{n}{k} a^{n-k} b^k \]
For our case, \(a = 3\), \(b = 2x\), and \(n = 50\). Therefore, the general term becomes:
\[ T_k = \binom{50}{k} (3)^{50-k} (2x)^k \]
Substituting \(x = \frac{1}{5}\):
\[ T_k = \binom{50}{k} (3)^{50-k} (2 \cdot \frac{1}{5})^k = \binom{50}{k} (3)^{50-k} \left(\frac{2}{5}\right)^k \]

### Step 2: Find the Ratio of Consecutive Terms
To find the longest term, we will analyze the ratio of consecutive terms \(T_k\) and \(T_{k+1}\):
\[ \frac{T_{k+1}}{T_k} = \frac{\binom{50}{k+1} (3)^{50-(k+1)} \left(\frac{2}{5}\right)^{k+1}}{\binom{50}{k} (3)^{50-k} \left(\frac{2}{5}\right)^k} \]
This simplifies to:
\[ \frac{T_{k+1}}{T_k} = \frac{\binom{50}{k+1}}{\binom{50}{k}} \cdot \frac{(3)^{50-(k+1)}}{(3)^{50-k}} \cdot \frac{\left(\frac{2}{5}\right)^{k+1}}{\left(\frac{2}{5}\right)^k} \]
\[ = \frac{50-k}{k+1} \cdot \frac{1}{3} \cdot \frac{2}{5} = \frac{(50-k) \cdot 2}{3 \cdot 5 \cdot (k+1)} = \frac{(50-k) \cdot 2}{15(k+1)} \]

### Step 3: Set the Ratio Equal to 1
To find the maximum term, we set the ratio equal to 1:
\[ \frac{(50-k) \cdot 2}{15(k+1)} = 1 \]
Cross-multiplying gives:
\[ (50-k) \cdot 2 = 15(k+1) \]
Expanding and rearranging:
\[ 100 - 2k = 15k + 15 \]
\[ 100 - 15 = 15k + 2k \]
\[ 85 = 17k \implies k = 5 \]

### Step 4: Determine the Longest Term
The longest term is either \(T_k\) or \(T_{k+1}\). Since \(k = 5\), we need to evaluate \(T_5\) and \(T_6\).

Calculating \(T_5\):
\[ T_5 = \binom{50}{5} (3)^{45} \left(\frac{2}{5}\right)^5 \]
Calculating \(T_6\):
\[ T_6 = \binom{50}{6} (3)^{44} \left(\frac{2}{5}\right)^6 \]

### Step 5: Compare \(T_5\) and \(T_6\)
To determine which is larger, we can compare the ratios:
\[ \frac{T_6}{T_5} = \frac{\binom{50}{6}}{\binom{50}{5}} \cdot \frac{(3)^{44}}{(3)^{45}} \cdot \frac{\left(\frac{2}{5}\right)^6}{\left(\frac{2}{5}\right)^5} \]
\[ = \frac{50-5}{6} \cdot \frac{1}{3} \cdot \frac{2}{5} = \frac{45 \cdot 2}{6 \cdot 15} = \frac{90}{90} = 1 \]
This indicates that \(T_5\) and \(T_6\) are equal, so both terms are numerically the longest.

### Final Answer
The numerically longest term in the expansion of \((3 + 2x)^{50}\) when \(x = \frac{1}{5}\) is both \(T_5\) and \(T_6\).