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If the coefficient of x^2 in the expansion of (4+kx) (2-x)^6 is 384,find the value of k and the coefficient of x^3 in the expansion?​

User Ricbit
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Answer:

The coefficient of x^2 in the expansion of (4+kx)(2-x)^6 is given by the product of the coefficients of x^2 in the two factors.

The coefficient of x^2 in (4+kx) is k times the coefficient of x in (4+kx), which is 4k.

The coefficient of x^2 in (2-x)^6 is given by the binomial coefficient (6,2) times (2)^4 times (-x)^2, which is 15*16x^2 or 240x^2.

Therefore, we have:

4k*240 = 384

Solving for k, we get:

k = 8/5

To find the coefficient of x^3 in the expansion, we use the same method.

The coefficient of x^3 in (4+kx)(2-x)^6 is given by the sum of the products of the coefficients of x^3 in the two factors.

The coefficient of x^3 in (4+kx) is k times the coefficient of x^3 in (4+kx), which is 4k.

The coefficient of x^3 in (2-x)^6 is given by the binomial coefficient (6,3) times (2)^3 times (-x)^3, which is -160x^3.

Therefore, we have:

4k*(-160) = -640k

Substituting k = 8/5, we get:

-640(8/5) = -1024

Therefore, the coefficient of x^3 in the expansion is -1024.

User ZalewaPL
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