Write the first three terms in the expansion of `(2-(y)/(3))^(6)`

Answer

To find the first three terms in the expansion of \( (2 - \frac{y}{3})^6 \), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

In our case, we have:
- \( a = 2 \)
- \( b = -\frac{y}{3} \)
- \( n = 6 \)

### Step 1: Identify the first three terms
We need to find the first three terms of the expansion, which correspond to \( k = 0, 1, 2 \).

### Step 2: Calculate the first term (\( k = 0 \))
Using the formula for the first term:

\[ \text{Term 1} = \binom{6}{0} (2)^{6} \left(-\frac{y}{3}\right)^{0} \]

Calculating this:

\[ \text{Term 1} = 1 \cdot 2^6 \cdot 1 = 64 \]

### Step 3: Calculate the second term (\( k = 1 \))
Using the formula for the second term:

\[ \text{Term 2} = \binom{6}{1} (2)^{6-1} \left(-\frac{y}{3}\right)^{1} \]

Calculating this:

\[ \text{Term 2} = 6 \cdot 2^5 \cdot \left(-\frac{y}{3}\right) = 6 \cdot 32 \cdot \left(-\frac{y}{3}\right) = -64y \]

### Step 4: Calculate the third term (\( k = 2 \))
Using the formula for the third term:

\[ \text{Term 3} = \binom{6}{2} (2)^{6-2} \left(-\frac{y}{3}\right)^{2} \]

Calculating this:

\[ \text{Term 3} = \frac{6 \cdot 5}{2 \cdot 1} \cdot 2^4 \cdot \left(-\frac{y}{3}\right)^{2} = 15 \cdot 16 \cdot \frac{y^2}{9} = \frac{240y^2}{9} = \frac{80y^2}{3} \]

### Step 5: Combine the terms
Now, we can combine the first three terms:

\[ (2 - \frac{y}{3})^6 \approx 64 - 64y + \frac{80y^2}{3} \]

### Final Result
The first three terms in the expansion of \( (2 - \frac{y}{3})^6 \) are:

\[ 64 - 64y + \frac{80y^2}{3} \]

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