Give the first 3 terms of the equation: (2p-y)52

To find the first three terms of the expansion of the binomial expression \((2p - y)^{52}\), we will use the Binomial Theorem. The Binomial Theorem states:

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

For our expression \((2p - y)^{52}\):

- \(a = 2p\)
- \(b = -y\)
- \(n = 52\)

The first three terms in the expansion are for \(k = 0\), \(k = 1\), and \(k = 2\).

### Finding the first three terms:

1. **First term** (\(k = 0\)):
   \[
   T_0 = \binom{52}{0} (2p)^{52} (-y)^0 = 1 \cdot (2p)^{52} \cdot 1 = (2p)^{52}
   \]

2. **Second term** (\(k = 1\)):
   \[
   T_1 = \binom{52}{1} (2p)^{51} (-y)^1 = 52 \cdot (2p)^{51} \cdot (-y) = -52 \cdot (2p)^{51} y
   \]

3. **Third term** (\(k = 2\)):
   \[
   T_2 = \binom{52}{2} (2p)^{50} (-y)^2 = \frac{52 \times 51}{2} \cdot (2p)^{50} \cdot y^2
   \]

   Simplifying \(T_2\):
   \[
   T_2 = 1326 \cdot (2p)^{50} \cdot y^2
   \]

Now, compute the terms:

\[
(2p)^{52}, \quad -52 \cdot (2p)^{51} y, \quad 1326 \cdot (2p)^{50} y^2
\]

### Final first three terms:

1. \( (2p)^{52} \)
2. \( -52 \cdot (2p)^{51} y \)
3. \( 1326 \cdot (2p)^{50} y^2 \)

Would you like more details or have any questions about this? Here are some related questions for further exploration:

1. How would you find the fourth term of the binomial expansion?
2. How does the Binomial Theorem work for negative exponents?
3. What are binomial coefficients, and how do you calculate them?
4. What is Pascal's Triangle, and how does it relate to binomial expansions?
5. How do we expand expressions involving more than two terms, like a trinomial?

**Tip:** The Binomial Theorem is a powerful tool for expanding expressions raised to a power, especially when calculating specific terms in large expansions.

Explore the binomial expansion of (2p - y)^52 and discover its first three terms using the Binomial Theorem. This problem covers advanced algebraic concepts and the application of the Binomial Theorem in expanding binomial expressions.

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