The coefficient of the middle term in the expansion of `(1+x)^40` is

Options

A. `(1.3.5----39)/(20!).2^20`
B. `(1.3.5---39)/(20!)`
C. `(40!)/(20!)`
D. `40!2^20`

Answer

To find the coefficient of the middle term in the expansion of \((1+x)^{40}\), we can follow these steps:

### Step 1: Identify the number of terms in the expansion
The expansion of \((1+x)^n\) has \(n+1\) terms. Here, \(n = 40\), so the number of terms is:
\[ 40 + 1 = 41 \]

**Hint:** Remember that the number of terms in the expansion is always \(n + 1\).

### Step 2: Determine the middle term
Since there are 41 terms (which is odd), the middle term is the \(\left(\frac{n}{2} + 1\right)\)th term. Thus, the middle term is:
\[ \frac{40}{2} + 1 = 20 + 1 = 21 \]

**Hint:** For an odd number of terms, the middle term is located at position \(\frac{n}{2} + 1\).

### Step 3: Write the expression for the 21st term
The general term in the binomial expansion of \((1+x)^n\) is given by:
\[ T_k = \binom{n}{k-1} x^{k-1} \]
For the 21st term, \(k = 21\), so:
\[ T_{21} = \binom{40}{20} x^{20} \]

**Hint:** The term number \(k\) corresponds to \(k-1\) in the binomial coefficient.

### Step 4: Find the coefficient of the middle term
The coefficient of \(x^{20}\) in the 21st term is \(\binom{40}{20}\). We need to calculate \(\binom{40}{20}\):
\[ \binom{40}{20} = \frac{40!}{20! \cdot 20!} \]

**Hint:** The binomial coefficient \(\binom{n}{r}\) can be calculated using the factorial formula.

### Step 5: Simplify the expression
To calculate \(\binom{40}{20}\), we can simplify it:
\[ \binom{40}{20} = \frac{40 \times 39 \times 38 \times \ldots \times 21}{20 \times 19 \times 18 \times \ldots \times 1} \]
This can also be expressed as:
\[ \binom{40}{20} = \frac{40!}{(20!)^2} \]

**Hint:** Factorials can be cumbersome, so consider canceling terms where possible.

### Step 6: Conclusion
The coefficient of the middle term in the expansion of \((1+x)^{40}\) is \(\binom{40}{20}\).

Thus, the final answer is:
\[ \text{Coefficient of the middle term} = \binom{40}{20} \]