Write the coefficient of the middle term in the expansion of `{(x+y^3)^3}^7dot`

Options

A. `(1*3*5....(2n-1))/(n!) 2^(n)`
B. `(1*3*5....(2n-1))/((n!)2^(n) )2^(n)`
C. `((2n)!)/((n!)2^(n) )2^(n)`
D. none of these

Answer

To find the coefficient of the middle term in the expansion of \((x + y^3)^{21}\), we will follow these steps:

### Step 1: Identify the total number of terms in the expansion
The expression \((x + y^3)^{21}\) can be expanded using the Binomial Theorem, which states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

Here, \(n = 21\).

### Step 2: Determine the number of terms
The total number of terms in the expansion is \(n + 1 = 21 + 1 = 22\).

### Step 3: Identify the middle term
Since the total number of terms (22) is even, there are two middle terms. The middle terms are the 11th and 12th terms in the expansion.

### Step 4: Write the general term
The general term \(T_r\) in the expansion is given by:

\[ T_r = \binom{21}{r} x^{21 - r} (y^3)^r = \binom{21}{r} x^{21 - r} y^{3r} \]

### Step 5: Find the 11th and 12th terms
- For the 11th term (\(r = 10\)):
\[ T_{10} = \binom{21}{10} x^{21 - 10} y^{30} = \binom{21}{10} x^{11} y^{30} \]

- For the 12th term (\(r = 11\)):
\[ T_{11} = \binom{21}{11} x^{21 - 11} y^{33} = \binom{21}{11} x^{10} y^{33} \]

### Step 6: Coefficients of the middle terms
The coefficients of the middle terms are:
- Coefficient of the 11th term: \(\binom{21}{10}\)
- Coefficient of the 12th term: \(\binom{21}{11}\)

### Step 7: Final answer
The coefficients of the middle terms are \(\binom{21}{10}\) and \(\binom{21}{11}\).