Find the middle term in the expansion of `(1+2x+x^(2))^(10)`

Explanation

To find the middle term in the expansion of \((1 + 2x + x^2)^{10}\), we can follow these steps:

### Step 1: Simplify the Expression
First, we can rewrite the expression \(1 + 2x + x^2\) as a perfect square:
\[ 1 + 2x + x^2 = (1 + x)^2 \]
Thus, we can express the original expression as:
\[ (1 + 2x + x^2)^{10} = ((1 + x)^2)^{10} = (1 + x)^{20} \]

### Step 2: Determine the Number of Terms
In the expansion of \((1 + x)^{20}\), the total number of terms is given by \(n + 1\), where \(n\) is the exponent. Here, \(n = 20\), so the total number of terms is:
\[ 20 + 1 = 21 \]

### Step 3: Identify the Middle Term
Since there are 21 terms, the middle term will be the \(11^{th}\) term (as the middle term is given by \(\frac{n + 1}{2}\)).

### Step 4: Use the Binomial Theorem
The \(r^{th}\) term in the binomial expansion is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(n = 20\), \(a = 1\), and \(b = x\). For the \(11^{th}\) term, we need \(r = 10\):
\[ T_{11} = \binom{20}{10} (1)^{20-10} (x)^{10} = \binom{20}{10} x^{10} \]

### Step 5: Calculate \(\binom{20}{10}\)
Now we need to calculate \(\binom{20}{10}\):
\[ \binom{20}{10} = \frac{20!}{10! \cdot 10!} \]
Calculating this gives us:
\[ \binom{20}{10} = 184756 \]

### Step 6: Write the Middle Term
Thus, the middle term \(T_{11}\) is:
\[ T_{11} = 184756 x^{10} \]

### Final Answer
The middle term in the expansion of \((1 + 2x + x^2)^{10}\) is:
\[ 184756 x^{10} \]