Use paseaf's triangle to sofve \( (2 x-1)^{5} \)

We can expand (2x – 1)^5 using the binomial theorem, which relies on Pascal’s triangle to give the binomial coefficients. For n = 5, the coefficients from Pascal’s triangle (the 6 numbers in the 5th row) are:   1 5 10 10 5 1 The binomial theorem states that   (a + b)^n = Σ from k = 0 to n of [nCk · a^(n–k) · b^k]. Here, let a = 2x and b = –1. Thus,   (2x – 1)^5 = Σ from k = 0 to 5 of [C(5, k) · (2x)^(5–k) · (–1)^k]. Now, let’s compute each term: 1. For k = 0:   Coefficient: C(5, 0) = 1   Term: 1 · (2x)^5 · (–1)^0 = 32x^5 2. For k = 1:   Coefficient: C(5, 1) = 5   Term: 5 · (2x)^4 · (–1)^1 = 5 · 16x^4 · (–1) = –80x^4 3. For k = 2:   Coefficient: C(5, 2) = 10   Term: 10 · (2x)^3 · (–1)^2 = 10 · 8x^3 · 1 = 80x^3 4. For k = 3:   Coefficient: C(5, 3) = 10   Term: 10 · (2x)^2 · (–1)^3 = 10 · 4x^2 · (–1) = –40x^2 5. For k = 4:   Coefficient: C(5, 4) = 5   Term: 5 · (2x)^1 · (–1)^4 = 5 · 2x · 1 = 10x 6. For k = 5:   Coefficient: C(5, 5) = 1   Term: 1 · (2x)^0 · (–1)^5 = 1 · 1 · (–1) = –1 Putting it all together, we have:   (2x – 1)^5 = 32x^5 – 80x^4 + 80x^3 – 40x^2 + 10x – 1