To solve these exercises using Pascal's Triangle, we'll expand each binomial. Pascal's Triangle provides the coefficients for the expansion.

Key Steps:

1. Identify the degree (n) of the binomial.
2. Use Pascal's Triangle to determine the coefficients.
3. Expand using the binomial theorem: $(a + b)^n = \sum_{k=0}^n C(n,k) \cdot a^{n-k} \cdot b^k$.

Let’s expand each one:

43. $(2t + 4)^3$

From Pascal's Triangle for $n = 3$, coefficients are $1, 3, 3, 1$. $(2t + 4)^3 = 1(2t)^3 + 3(2t)^2(4) + 3(2t)(4)^2 + 1(4)^3$ $= 8t^3 + 48t^2 + 96t + 64$

44. $(6m + 2)^2$

For $n = 2$, coefficients are $1, 2, 1$. $(6m + 2)^2 = 1(6m)^2 + 2(6m)(2) + 1(2)^2$ $= 36m^2 + 24m + 4$

45. $(2q - 3)^4$

For $n = 4$, coefficients are $1, 4, 6, 4, 1$. $(2q - 3)^4 = 1(2q)^4 + 4(2q)^3(-3) + 6(2q)^2(-3)^2 + 4(2q)(-3)^3 + 1(-3)^4$ $= 16q^4 - 96q^3 + 216q^2 - 216q + 81$

46. $(g + 2)^5$

For $n = 5$, coefficients are $1, 5, 10, 10, 5, 1$. $(g + 2)^5 = 1g^5 + 5g^4(2) + 10g^3(2)^2 + 10g^2(2)^3 + 5g(2)^4 + 1(2)^5$ $= g^5 + 10g^4 + 40g^3 + 80g^2 + 80g + 32$

47. $(yz + 1)^5$

Using the same coefficients for $n = 5$ as above: $(yz + 1)^5 = 1(yz)^5 + 5(yz)^4(1) + 10(yz)^3(1)^2 + 10(yz)^2(1)^3 + 5(yz)(1)^4 + 1(1)^5$ $= (yz)^5 + 5(yz)^4 + 10(yz)^3 + 10(yz)^2 + 5(yz) + 1$

48. $(np - 1)^4$

For $n = 4$, coefficients are $1, 4, 6, 4, 1$: $(np - 1)^4 = 1(np)^4 + 4(np)^3(-1) + 6(np)^2(-1)^2 + 4(np)(-1)^3 + 1(-1)^4$ $= (np)^4 - 4(np)^3 + 6(np)^2 - 4(np) + 1$

Let me know if you'd like detailed steps for any of the expansions or explanations on Pascal's Triangle.

Related Questions:

1. How do you derive coefficients for Pascal's Triangle?
2. How is the binomial theorem connected to Pascal's Triangle?
3. Can Pascal's Triangle be used for expansions with negative or fractional exponents?
4. How does the order of the terms affect the expansion of a binomial?
5. How would the expansion differ for a trinomial like $(a + b + c)^n$?

Tip:

Remember that Pascal’s Triangle is symmetric, so you can use this property to quickly find coefficients!