Question

1. which is the correct expansion of \\((x + 4y)^3\\) using the binomial theorem?

\\((x + 4y)^3 = (x)^3(4y)^0 + (x)^2(4y)^1 + (x)^2(4y)^2 + (x)^3(4y)^3\\)

\\((x + 4y)^3 = (4y)^0 + (x)^2(4y)^1 + (x)^1(4y)^2 + (x)^0(4y)^3\\)

\\((x + 4y)^3 = \binom{3}{0}(x)^3(4y)^0 + \binom{3}{1}(x)^2(4y)^1 + \binom{3}{2}(x)^1(4y)^2 + \binom{3}{3}(x)^0(4y)^3\\)

\\((x + 4y)^3 = \binom{3}{0}(4y)^3(x)^0 + \binom{3}{1}(4y)^2(x)^1 + \binom{3}{2}(4y)^1(x)^2 + \binom{3}{3}(4y)^0(x)^3\\)

Explanation:

🆕 New Concept Discovered: Binomial Theorem
Expanding binomials using combinations and powers.

Step 1: Understand the Binomial Theorem Formula

The Binomial Theorem states that for any positive integer \(n\):

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

For our expression, we have:

• \(a = x\)
• \(b = 4y\)
• \(n = 3\)

This means the expansion will have 4 terms, where \(k\) goes from \(0\) to \(3\).

Step 2: Set up the expansion terms

Using the formula, we write out each term for \(k = 0, 1, 2, 3\):

• For \(k = 0\):

$$ \binom{3}{0} (x)^{3-0} (4y)^0 = \binom{3}{0} (x)^3 (4y)^0 $$

• For \(k = 1\):

$$ \binom{3}{1} (x)^{3-1} (4y)^1 = \binom{3}{1} (x)^2 (4y)^1 $$

• For \(k = 2\):

$$ \binom{3}{2} (x)^{3-2} (4y)^2 = \binom{3}{2} (x)^1 (4y)^2 $$

• For \(k = 3\):

$$ \binom{3}{3} (x)^{3-3} (4y)^3 = \binom{3}{3} (x)^0 (4y)^3 $$

Step 3: Combine the terms and match the options

Putting all the terms together:

$$ (x + 4y)^3 = \binom{3}{0} (x)^3 (4y)^0 + \binom{3}{1} (x)^2 (4y)^1 + \binom{3}{2} (x)^1 (4y)^2 + \binom{3}{3} (x)^0 (4y)^3 $$

Comparing this with the given choices, it matches the third option exactly.

Answer:

The correct option is the third one:

$$ (x + 4y)^3 = \binom{3}{0}(x)^3(4y)^0 + \binom{3}{1}(x)^2(4y)^1 + \binom{3}{2}(x)^1(4y)^2 + \binom{3}{3}(x)^0(4y)^3 $$