x^2(3x^2+k/x)^8 The constant is 16128. Find k

Question

 This is binomial expansion. I need to find k. The constant is 16128.

 

x²(3x²+k/x)⁸

Answer 1

Let's figure out how to find the constant term. The binomial part (3x² + k/x)⁸ will be expanded like this, where a_i are the coefficients in the 8th row in Pascal's Triangle (1 8 28 56 70 56 28 8 1):

 

(3x²)⁸ + a₂(3x²)⁷(k/x) + a₃(3x²)⁶(k/x)² + a₄(3x²)⁵(k/x)³ + a₅(3x²)⁴(k/x)⁴ + a₆(3x²)³(k/x)⁵ + a₇(3x²)²(k/x)⁶ + a₈(3x²)(k/x)⁷ + (k/x)⁸

 

 

 

We'd like to find the x^-2 term, since multiplying that by x² will result in a constant.

 

The first term will have x⁽²⁾⁽⁸⁾ = x¹⁶. The second will have x^(2)(7)-1 = x¹³. The third will have x^{(2)(6)-(1)(2)} = x¹⁰. The fourth will have x^{(2)(5)-(1)(3)} = x⁷. The fifth will have x^{(2)(4)-(1)(4)} = x⁴.

 

The exponent seems to be going down by 3 each time, so the sixth term will be x¹ and the seventh term will be x^-2, like we want.

 

 

 

The seventh term is:

 

a₇(3x²)²(k/x)⁶ = a₇(9x⁴)(k⁶/x⁶) = a₇(9k⁶/x²) = 28(9k⁶)/x² = 252k⁶/x²

 

 

 

Multiply that by x² to get 252k⁶, our constant term. So:

 

252k⁶ = 16128

k⁶ = 64

k = 2