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: Identify the structure of the expression
The expression can be rewritten as:
\[ (1 + 2x + x^2)^{10} \]
This is a trinomial expansion, but we can also treat it as a binomial expansion by recognizing that \(1 + 2x + x^2\) can be expressed in a different way.

### Step 2: Rewrite the expression
We can rewrite \(1 + 2x + x^2\) as:
\[ (1 + x)^2 \]
Thus, we have:
\[ (1 + 2x + x^2)^{10} = ((1 + x)^2)^{10} = (1 + x)^{20} \]

### Step 3: Determine the number of terms
In the expansion of \((1 + x)^{20}\), the number of terms is \(n + 1\), where \(n\) is the exponent. Here, \(n = 20\), so the number of terms is:
\[ 20 + 1 = 21 \]

### Step 4: Find the middle term
Since the number of terms is odd (21), the middle term is the \(\frac{21 + 1}{2} = 11\)th term.

### Step 5: Use the binomial theorem to find the 11th term
The general term in the expansion of \((1 + x)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} \cdot (1)^{n-r} \cdot (x)^r \]
For the 11th term (\(T_{11}\)), we have \(r = 10\) (since we start counting from \(T_1\)):
\[ T_{11} = \binom{20}{10} \cdot (1)^{20-10} \cdot (x)^{10} = \binom{20}{10} \cdot x^{10} \]

### Step 6: Calculate \(\binom{20}{10}\)
Using the combination formula:
\[ \binom{20}{10} = \frac{20!}{10! \cdot 10!} = 184756 \]

### Step 7: Write the final middle term
Thus, the middle term is:
\[ T_{11} = 184756 \cdot x^{10} \]

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

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