Expand `(1+ 2 x + 3x^(2) )^(n)` in a series of ascending powers of x up to and including the term in `x^2`.

Answer

To expand the expression \((1 + 2x + 3x^2)^n\) in a series of ascending powers of \(x\) up to and including the term in \(x^2\), we can follow these steps:

### Step 1: Identify the Binomial Expansion
We will use the binomial theorem, which states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, we can treat \(a = 1\) and \(b = 2x + 3x^2\).

### Step 2: Expand the Expression
We can write:
\[ (1 + (2x + 3x^2))^n \]
Now, we will expand this using the binomial theorem:
\[ (1 + (2x + 3x^2))^n = \sum_{k=0}^{n} \binom{n}{k} (2x + 3x^2)^k \]

### Step 3: Calculate Terms for \(k = 0, 1, 2\)
We only need to consider terms up to \(k = 2\) since we want the expansion up to \(x^2\).

1. **For \(k = 0\)**:
\[ \binom{n}{0} (2x + 3x^2)^0 = 1 \]

2. **For \(k = 1\)**:
\[ \binom{n}{1} (2x + 3x^2)^1 = n(2x + 3x^2) = 2nx + 3nx^2 \]

3. **For \(k = 2\)**:
\[ \binom{n}{2} (2x + 3x^2)^2 = \frac{n(n-1)}{2} (2x + 3x^2)^2 \]
Now, we expand \((2x + 3x^2)^2\):
\[ (2x + 3x^2)^2 = 4x^2 + 12x^3 + 9x^4 \]
However, since we only need terms up to \(x^2\), we take:
\[ \frac{n(n-1)}{2} \cdot 4x^2 = 2n(n-1)x^2 \]

### Step 4: Combine All Terms
Now, we combine all the terms we calculated:
\[ 1 + (2nx + 3nx^2) + (2n(n-1)x^2) \]
This simplifies to:
\[ 1 + 2nx + (3n + 2n(n-1))x^2 \]
Combining the coefficients of \(x^2\):
\[ 3n + 2n(n-1) = 3n + 2n^2 - 2n = 2n^2 + n \]
Thus, the final expansion up to \(x^2\) is:
\[ 1 + 2nx + (2n^2 + n)x^2 \]

### Final Answer
The expansion of \((1 + 2x + 3x^2)^n\) up to and including the term in \(x^2\) is:
\[ 1 + 2nx + (2n^2 + n)x^2 \]