The coefficient of `x^(4)` in the expansion of `(1+x+x^(2))^(6)` is

Options

A. 72
B. 90
C. 96
D. 112

Answer

To find the coefficient of \(x^4\) in the expansion of \((1 + x + x^2)^6\), we can use the multinomial expansion. Here are the steps to solve the problem:

### Step 1: Understand the expression
The expression we need to expand is \((1 + x + x^2)^6\). We want to find the coefficient of \(x^4\) in this expansion.

### Step 2: Use the multinomial theorem
According to the multinomial theorem, the expansion of \((a_1 + a_2 + a_3)^n\) can be expressed as:
\[ \sum \frac{n!}{k_1! k_2! k_3!} a_1^{k_1} a_2^{k_2} a_3^{k_3} \]
where \(k_1 + k_2 + k_3 = n\).

In our case, \(a_1 = 1\), \(a_2 = x\), \(a_3 = x^2\), and \(n = 6\).

### Step 3: Identify the terms contributing to \(x^4\)
We need to find combinations of \(k_1\), \(k_2\), and \(k_3\) such that:
\[ k_2 + 2k_3 = 4 \]
and
\[ k_1 + k_2 + k_3 = 6 \]

### Step 4: Solve the equations
From the second equation, we can express \(k_1\) in terms of \(k_2\) and \(k_3\):
\[ k_1 = 6 - k_2 - k_3 \]

Substituting \(k_1\) into the first equation:
\[ k_2 + 2k_3 = 4 \]

Now we can find possible values for \(k_2\) and \(k_3\):

1. **If \(k_3 = 0\)**:
\[ k_2 = 4 \quad \Rightarrow \quad k_1 = 6 - 4 - 0 = 2 \]
Contribution: \(\frac{6!}{2!4!0!} = 15\)

2. **If \(k_3 = 1\)**:
\[ k_2 + 2(1) = 4 \quad \Rightarrow \quad k_2 = 2 \quad \Rightarrow \quad k_1 = 6 - 2 - 1 = 3 \]
Contribution: \(\frac{6!}{3!2!1!} = 60\)

3. **If \(k_3 = 2\)**:
\[ k_2 + 2(2) = 4 \quad \Rightarrow \quad k_2 = 0 \quad \Rightarrow \quad k_1 = 6 - 0 - 2 = 4 \]
Contribution: \(\frac{6!}{4!0!2!} = 15\)

### Step 5: Sum the contributions
Now, we sum up all the contributions:
\[ 15 + 60 + 15 = 90 \]

### Final Answer
The coefficient of \(x^4\) in the expansion of \((1 + x + x^2)^6\) is **90**.

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