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 will 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, \(a = 2\), \(b = -\frac{y}{3}\), and \(n = 6\).

### Step 1: Identify the terms for the expansion
Using the Binomial Theorem, we can express the first three terms of the expansion as follows:

1. For \(k = 0\):
\[ \binom{6}{0} (2)^{6} \left(-\frac{y}{3}\right)^{0} = 1 \cdot 64 \cdot 1 = 64 \]

2. For \(k = 1\):
\[ \binom{6}{1} (2)^{5} \left(-\frac{y}{3}\right)^{1} = 6 \cdot 32 \cdot \left(-\frac{y}{3}\right) = -\frac{192y}{3} = -64y \]

3. For \(k = 2\):
\[ \binom{6}{2} (2)^{4} \left(-\frac{y}{3}\right)^{2} = 15 \cdot 16 \cdot \left(\frac{y^2}{9}\right) = \frac{240y^2}{9} = \frac{80y^2}{3} \]

### Step 2: Combine the terms
Now, we can write the first three terms of the expansion together:

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

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

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