Verified answer

1) tem n+1 termos(4+1=5), o médio é t3......p+1=3>>p=2

t(p+1)=C(n,p).a^(n-p).b^p

t(2+1)=C(4,2).(x)².(-1)²

t3=4.3/2! . x² . 1

t3=6x² >>>>>

2)

(2+3)^6=5^6= 15625 >>>

binômio de Newton

(a + b)^4 = [1, 4, 6, 4, 1]

(a + b)^6 = [1, 6, 15, 20, 15, 6, 1]

(a + b)^9 = [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]

1) (x - 1)^4 = 6x²

2)

(2x + 3y)^6 =

(2x^6)*(3y^0) + (2x^5).(3y)^1 + (2x)^4(3y)^2 + (2x)^3(3y)^3 +

(2x)^2*(3y)^4 + (2x)^1*(3y)^5 + (2x)^0*(3y)^6 =

64x^6 + 32*x^5.3y + 16.x^4.9y^2 + 8x^3.27y^3 + 4x^2.81y^4 + 2x.243y^2 + 729y^6

soma = 64 + 96 + 144 + 216 + 324 + 486 + 729 = 2059

3) (a + b)^9 = [c1,c2,c3,c4,c5,c6,c7,c8,c9,c10] = [1,9,36,84,126,126,84,36,9,1]

(1/x + x¹/²)^9 =

c1/x^9 + c2.x^1/2/x^8 + c3.x^1/x^7 + c4.x^3/2/x^6 + c5.x^2/x^5 +

c6.x^5/2/x^4 + c7.x^3/x^3 + c8.x^7/2/x^2 + c9.x^4/x + c10.x^9/2

o termo independente é c7.x^3/x^3 = 84

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