The coefficient of middle term in the expansion of `(1+x)^(10)` is

Options

A. `10!//5!6!`
B. `10!//5!^(2)`
C. `10!//5!.7!`
D. None of these

Answer

To find the coefficient of the middle term in the expansion of \((1+x)^{10}\), we can follow these steps:

### Step 1: Identify the total number of terms
In the expansion of \((1+x)^n\), the total number of terms is \(n + 1\). Here, \(n = 10\), so the total number of terms is:
\[ 10 + 1 = 11 \]

### Step 2: Determine the middle term
Since the total number of terms is odd, the middle term will be the \((\frac{n}{2} + 1)\)th term. For \(n = 10\):
\[ \text{Middle term} = \left(\frac{10}{2} + 1\right) = 6 \text{th term} \]

### Step 3: Use the binomial theorem to find the term
The general term in the expansion of \((1+x)^n\) is given by:
\[ T_k = \binom{n}{k-1} x^{k-1} \]
For the 6th term, \(k = 6\):
\[ T_6 = \binom{10}{6-1} x^{6-1} = \binom{10}{5} x^5 \]

### Step 4: Calculate the coefficient
The coefficient of the middle term is given by \(\binom{10}{5}\). We can calculate this using the formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Substituting \(n = 10\) and \(r = 5\):
\[ \binom{10}{5} = \frac{10!}{5! \cdot (10-5)!} = \frac{10!}{5! \cdot 5!} \]

### Step 5: Simplify the expression
Calculating \(10!\) and \(5!\):
\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5! \]
Thus,
\[ \binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \]
Calculating the numerator:
\[ 10 \times 9 = 90 \]
\[ 90 \times 8 = 720 \]
\[ 720 \times 7 = 5040 \]
\[ 5040 \times 6 = 30240 \]
Now calculating the denominator:
\[ 5! = 120 \]
So,
\[ \binom{10}{5} = \frac{30240}{120} = 252 \]

### Final Answer
The coefficient of the middle term in the expansion of \((1+x)^{10}\) is \(252\).

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