Greatest term in the binomial expansion of `(a + 2x)^9` when `a=1 & x =1/3` is :

Options

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

Answer

To find the greatest term in the binomial expansion of \((a + 2x)^9\) when \(a = 1\) and \(x = \frac{1}{3}\), we can follow these steps:

### Step 1: Substitute the values of \(a\) and \(x\)
We start by substituting \(a = 1\) and \(x = \frac{1}{3}\) into the expression:
\[ (1 + 2 \cdot \frac{1}{3})^9 = (1 + \frac{2}{3})^9 = \left(\frac{5}{3}\right)^9 \]

### Step 2: Identify the general term in the binomial expansion
The general term \(T_k\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_k = \binom{n}{k} a^{n-k} b^k \]
For our case, \(n = 9\), \(a = 1\), and \(b = 2x = \frac{2}{3}\). Thus, the general term becomes:
\[ T_k = \binom{9}{k} (1)^{9-k} \left(\frac{2}{3}\right)^k = \binom{9}{k} \left(\frac{2}{3}\right)^k \]

### Step 3: Find the term \(T_k\) that is greatest
To find the greatest term, we can use the ratio of consecutive terms:
\[ \frac{T_{k+1}}{T_k} = \frac{\binom{9}{k+1} \left(\frac{2}{3}\right)^{k+1}}{\binom{9}{k} \left(\frac{2}{3}\right)^k} = \frac{\binom{9}{k+1}}{\binom{9}{k}} \cdot \frac{2}{3} \]
Using the property of binomial coefficients:
\[ \frac{\binom{9}{k+1}}{\binom{9}{k}} = \frac{9-k}{k+1} \]
Thus, we have:
\[ \frac{T_{k+1}}{T_k} = \frac{9-k}{k+1} \cdot \frac{2}{3} \]

### Step 4: Set the ratio to 1 to find the maximum term
To find the maximum term, we set:
\[ \frac{9-k}{k+1} \cdot \frac{2}{3} = 1 \]
Solving this gives:
\[ \frac{9-k}{k+1} = \frac{3}{2} \]
Cross-multiplying:
\[ 2(9-k) = 3(k+1) \]
Expanding:
\[ 18 - 2k = 3k + 3 \]
Combining like terms:
\[ 18 - 3 = 3k + 2k \implies 15 = 5k \implies k = 3 \]

### Step 5: Calculate the greatest term \(T_k\)
Now, we find the term \(T_3\):
\[ T_3 = \binom{9}{3} \left(\frac{2}{3}\right)^3 \]
Calculating \(\binom{9}{3}\):
\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]
Now substituting back:
\[ T_3 = 84 \cdot \left(\frac{2}{3}\right)^3 = 84 \cdot \frac{8}{27} = \frac{672}{27} \]

### Final Answer
Thus, the greatest term in the binomial expansion of \((1 + 2 \cdot \frac{1}{3})^9\) is:
\[ \frac{672}{27} \]