The numerically greatest term in the expansion of `(1 + x)^(10)`
when ` x = 2//3`, is

Options

A. `4^(th)`
B. `5^(th)`
C. `6^(th) `
D. `3^(rd)`

Explanation

To find the numerically greatest term in the expansion of \((1 + x)^{10}\) when \(x = \frac{2}{3}\), we will follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \(T_r\) in the expansion of \((1 + x)^n\) is given by:
\[ T_r = \binom{n}{r} x^r \]
For our case, \(n = 10\) and \(x = \frac{2}{3}\). Thus, the general term becomes:
\[ T_r = \binom{10}{r} \left(\frac{2}{3}\right)^r \]

### Step 2: Find the ratio of consecutive terms
To find the term that is numerically greatest, we will consider the ratio of consecutive terms \(T_{r+1}\) and \(T_r\):
\[ \frac{T_{r+1}}{T_r} = \frac{\binom{10}{r+1} \left(\frac{2}{3}\right)^{r+1}}{\binom{10}{r} \left(\frac{2}{3}\right)^r} \]
This simplifies to:
\[ \frac{T_{r+1}}{T_r} = \frac{\binom{10}{r+1}}{\binom{10}{r}} \cdot \frac{2}{3} \]

### Step 3: Simplify the ratio
Using the property of binomial coefficients:
\[ \frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k} \]
we can write:
\[ \frac{T_{r+1}}{T_r} = \frac{10 - r}{r + 1} \cdot \frac{2}{3} \]

### Step 4: Set the ratio greater than 1
To find the maximum term, we set the ratio greater than 1:
\[ \frac{10 - r}{r + 1} \cdot \frac{2}{3} > 1 \]
This leads to:
\[ \frac{10 - r}{r + 1} > \frac{3}{2} \]

### Step 5: Cross-multiply and solve the inequality
Cross-multiplying gives:
\[ 2(10 - r) > 3(r + 1) \]
Expanding both sides:
\[ 20 - 2r > 3r + 3 \]
Combining like terms:
\[ 20 - 3 > 3r + 2r \]
\[ 17 > 5r \]
Thus:
\[ r < \frac{17}{5} = 3.4 \]

### Step 6: Determine the integer value of \(r\)
Since \(r\) must be an integer, the largest possible value for \(r\) is 3.

### Step 7: Identify the term
The term \(T_{r+1}\) will be the greatest term. Since \(r = 3\), we find:
\[ T_{4} = \binom{10}{4} \left(\frac{2}{3}\right)^4 \]

### Step 8: Calculate the term
Calculating \(T_4\):
\[ T_4 = \binom{10}{4} \left(\frac{2}{3}\right)^4 = \frac{10!}{4!(10-4)!} \cdot \left(\frac{2}{3}\right)^4 \]
Calculating \(\binom{10}{4}\):
\[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]
Thus:
\[ T_4 = 210 \cdot \left(\frac{2}{3}\right)^4 = 210 \cdot \frac{16}{81} = \frac{3360}{81} \]

### Final Answer
The numerically greatest term in the expansion of \((1 + x)^{10}\) when \(x = \frac{2}{3}\) is:
\[ \frac{3360}{81} \]