To expand $(3a + 2b)^7$ using Pascal's Triangle, we follow these steps:

1. Identify the coefficients using Pascal's Triangle for $n = 7$: The 7th row of Pascal's Triangle is: $1, 7, 21, 35, 35, 21, 7, 1$

2. Apply the binomial theorem: The binomial expansion for $(x + y)^n$ is given by: $(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$ For $(3a + 2b)^7$, $x = 3a$, $y = 2b$, and $n = 7$. Now, expand term by term:

$(3a + 2b)^7 = \sum_{k=0}^{7} \binom{7}{k} (3a)^{7-k} (2b)^k$

Using the coefficients from Pascal's Triangle, we get:

$= 1(3a)^7(2b)^0 + 7(3a)^6(2b)^1 + 21(3a)^5(2b)^2 + 35(3a)^4(2b)^3 + 35(3a)^3(2b)^4 + 21(3a)^2(2b)^5 + 7(3a)^1(2b)^6 + 1(3a)^0(2b)^7$

3. Simplify each term:

• First term: $(3a)^7 = 2187a^7$
• Second term: $7(3a)^6(2b) = 7 \cdot 729a^6 \cdot 2b = 10206a^6b$
• Third term: $21(3a)^5(2b)^2 = 21 \cdot 243a^5 \cdot 4b^2 = 20412a^5b^2$
• Fourth term: $35(3a)^4(2b)^3 = 35 \cdot 81a^4 \cdot 8b^3 = 22680a^4b^3$
• Fifth term: $35(3a)^3(2b)^4 = 35 \cdot 27a^3 \cdot 16b^4 = 15120a^3b^4$
• Sixth term: $21(3a)^2(2b)^5 = 21 \cdot 9a^2 \cdot 32b^5 = 6048a^2b^5$
• Seventh term: $7(3a)^1(2b)^6 = 7 \cdot 3a \cdot 64b^6 = 1344ab^6$
• Eighth term: $(3a)^0(2b)^7 = 128b^7$

4. Final expanded form:

$(3a + 2b)^7 = 2187a^7 + 10206a^6b + 20412a^5b^2 + 22680a^4b^3 + 15120a^3b^4 + 6048a^2b^5 + 1344ab^6 + 128b^7$

This is the complete expansion of $(3a + 2b)^7$.

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Would you like any clarifications or further details on this?

Here are 5 related questions to expand on this topic:

1. How would you expand $(4x + y)^6$ using Pascal's Triangle?
2. What is Pascal's Triangle, and how is it derived?
3. Can you find the expansion of $(2a - b)^5$ using the binomial theorem?
4. How does the binomial theorem apply to negative exponents?
5. How can Pascal's Triangle be used in combinatorics?

Tip: Pascal's Triangle can help solve combinatorial problems like finding the number of ways to choose items, as its entries correspond to binomial coefficients.