In the expansion of `(1/2 + 2x/3)^n` when `x=-1/2` ,it is known that `3^(rd)` term is the greatest term Find the possible integral values of n .

Answer

To solve the problem, we need to analyze the binomial expansion of \((\frac{1}{2} + \frac{2x}{3})^n\) when \(x = -\frac{1}{2}\). We are tasked with finding the integral values of \(n\) such that the third term is the greatest term in the expansion.

### Step 1: Substitute the value of \(x\)

Substituting \(x = -\frac{1}{2}\) into the expression gives us:

\[ \left(\frac{1}{2} + \frac{2(-\frac{1}{2})}{3}\right)^n = \left(\frac{1}{2} - \frac{1}{3}\right)^n = \left(\frac{3}{6} - \frac{2}{6}\right)^n = \left(\frac{1}{6}\right)^n \]

### Step 2: Identify the terms in the binomial expansion

The \(r\)-th term in the binomial expansion of \((a + b)^n\) is given by:

\[ T_r = \binom{n}{r-1} a^{n-(r-1)} b^{r-1} \]

In our case, \(a = \frac{1}{2}\) and \(b = \frac{2(-\frac{1}{2})}{3} = -\frac{1}{3}\).

Thus, the \(r\)-th term can be expressed as:

\[ T_r = \binom{n}{r-1} \left(\frac{1}{2}\right)^{n-(r-1)} \left(-\frac{1}{3}\right)^{r-1} \]

### Step 3: Find the third term \(T_3\) and the second term \(T_2\)

The third term \(T_3\) is:

\[ T_3 = \binom{n}{2} \left(\frac{1}{2}\right)^{n-2} \left(-\frac{1}{3}\right)^{2} \]

Calculating this gives:

\[ T_3 = \binom{n}{2} \left(\frac{1}{2}\right)^{n-2} \left(\frac{1}{9}\right) \]

The second term \(T_2\) is:

\[ T_2 = \binom{n}{1} \left(\frac{1}{2}\right)^{n-1} \left(-\frac{1}{3}\right)^{1} \]

Calculating this gives:

\[ T_2 = n \left(\frac{1}{2}\right)^{n-1} \left(-\frac{1}{3}\right) \]

### Step 4: Set up the inequality for the greatest term

Since \(T_3\) is the greatest term, we need to satisfy the following conditions:

1. \(T_3 \geq T_2\)
2. \(T_3 \geq T_4\)

First, we will compare \(T_3\) and \(T_2\):

\[ T_3 \geq T_2 \implies \binom{n}{2} \left(\frac{1}{2}\right)^{n-2} \cdot \frac{1}{9} \geq n \left(\frac{1}{2}\right)^{n-1} \cdot \left(-\frac{1}{3}\right) \]

### Step 5: Simplify the inequality

This simplifies to:

\[ \frac{n(n-1)}{2} \cdot \frac{1}{9} \geq n \cdot \frac{1}{2} \cdot \frac{1}{3} \]

Cancelling \(n\) (assuming \(n \neq 0\)) gives:

\[ \frac{(n-1)}{2} \cdot \frac{1}{9} \geq \frac{1}{6} \]

Multiplying through by 18 gives:

\[ (n-1) \geq 2 \implies n \geq 3 \]

### Step 6: Compare \(T_3\) and \(T_4\)

Next, we compare \(T_3\) and \(T_4\):

\[ T_4 \leq T_3 \implies \binom{n}{3} \left(\frac{1}{2}\right)^{n-3} \cdot \left(-\frac{1}{3}\right)^{3} \leq \binom{n}{2} \left(\frac{1}{2}\right)^{n-2} \cdot \left(-\frac{1}{3}\right)^{2} \]

This simplifies to:

\[ \frac{n(n-1)(n-2)}{6} \cdot \frac{1}{27} \leq \frac{n(n-1)}{2} \cdot \frac{1}{9} \]

Cancelling \(n(n-1)\) (assuming \(n > 1\)) gives:

\[ \frac{(n-2)}{6} \cdot \frac{1}{27} \leq \frac{1}{18} \]

Multiplying through by 162 gives:

\[ n-2 \leq 9 \implies n \leq 11 \]

### Conclusion: Integral values of \(n\)

Combining the inequalities \(3 \leq n \leq 11\), the possible integral values of \(n\) are:

\[ n = 3, 4, 5, 6, 7, 8, 9, 10, 11 \]