Find the term independent of x in the expansion of (2x+1-1/2x^2)^6

We are asked to find the term independent of \( x \) in the expansion of \( (2x + 1 - \frac{1}{2x^2})^6 \). This is a trinomial expansion.

### Step 1: Use the multinomial theorem
The multinomial theorem generalizes the binomial theorem for expressions with more than two terms. For a trinomial \( (a + b + c)^n \), the general term is given by:
\[
T = \frac{n!}{k_1! k_2! k_3!} a^{k_1} b^{k_2} c^{k_3}
\]
where \( k_1 + k_2 + k_3 = n \).

Here, \( a = 2x \), \( b = 1 \), and \( c = -\frac{1}{2x^2} \), and \( n = 6 \).

### Step 2: Analyze the powers of \( x \)
The goal is to find the term where the powers of \( x \) cancel out, i.e., where the overall power of \( x \) in the expansion is zero. 

- \( (2x)^{k_1} \) contributes a factor of \( 2^{k_1} x^{k_1} \).
- \( 1^{k_2} \) contributes no factor of \( x \).
- \( \left( -\frac{1}{2x^2} \right)^{k_3} \) contributes a factor of \( (-1)^{k_3} \left(\frac{1}{2}\right)^{k_3} x^{-2k_3} \).

The total power of \( x \) in a given term is:
\[
\text{Power of } x = k_1 - 2k_3.
\]
We want this power to be zero:
\[
k_1 - 2k_3 = 0 \quad \Rightarrow \quad k_1 = 2k_3.
\]

### Step 3: Solve for \( k_1, k_2, k_3 \)
Since \( k_1 + k_2 + k_3 = 6 \), we substitute \( k_1 = 2k_3 \) into the equation:
\[
2k_3 + k_2 + k_3 = 6 \quad \Rightarrow \quad 3k_3 + k_2 = 6.
\]
This gives possible values for \( k_3 \) and \( k_2 \):

- If \( k_3 = 0 \), then \( k_2 = 6 \), and \( k_1 = 0 \).
- If \( k_3 = 1 \), then \( k_2 = 3 \), and \( k_1 = 2 \).
- If \( k_3 = 2 \), then \( k_2 = 0 \), and \( k_1 = 4 \).

### Step 4: Find the

Learn how to find the term independent of x in the trinomial expansion of (2x + 1 - 1/2x^2)^6 using the multinomial theorem. This step-by-step guide explains the analysis of powers of x and provides solutions suitable for students in Grades 10-12.

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