Question

Quero saber como se desenvolve o binômio de newton: (2x^2-y^3)^4?

Answer 1

n
∑ (a+b)^n = a^(n-k)*b^(k)*Cn,k
k=0
                         4
(2x²+(-y³))⁴ = ∑  
                       k =0

(2x²)⁴⁻⁰*(-y³)⁰*C4,0 + (2x²)⁴⁻¹*(-y³)¹*C4,1 + (2x²)⁴⁻²*(-y³)²*C4,2 +(2x²)⁴⁻³*

(-y³)³*C4,3 +(2x²)⁴⁻⁴*(-y³)⁴*C4,4 =

↓

(2x²)⁴*C4,0 +(2x²)³*(-y³)¹*C4,1 +(2x²)²*(-y³)²*C4,2+(2x²)¹(-y³)³*C4,3+(2x²)⁰*(-y³)⁴*C,4,4

↓

Vamos calcular os Cn,k

$\\ Cn,k= \frac{n!}{(n-k)!k!} \\ \\ C(4,0) = \frac{4!}{0!(4-0)!} = 4!/4! = 1 \\ \\ C4,1 = \frac{4!}{1!3!} = \frac{4*3!}{3!} = 4 \\ \\ C4,2= \frac{4!}{2!2!} = \frac{4*3*2!}{2!*2*1} = \frac{4*3}{2} =6 \\ \\ C4,3 = \frac{4!}{3!1!} = \frac{4*3!}{3!} =4 \\ \\ C4,4= \frac{4!}{4!0!} = 1$

Substituindo ficamos:

16x⁸*1 +8x⁶*(-y³)*4 + 4x⁴*y⁸*6+2x²*(-y⁹)*4+1*y¹²*1

↓

16x⁸-32x⁶y³+24x⁴y⁸-8x²y⁹+y¹²