Working of coefficients

Probability Level 3

Find the coefficient of $x^3$ in the following expansion:

$\left(1+\frac{1}{2}x-3x^2\right)^{20}.$

1 solution

Sara Cart
Jun 29, 2015

** Note: some number C another number means a combination.

Set it up as (-3x^2 + (1 + 1/2x))^20 --> easier to see as binomial expansion

The possibilities for getting x^3 are: x^2 * x x^3 straight away

x^2 *x: The term of the binomial expansion needed is: 20 C 1 (-3x^2)^1 * (1 + 1/2x)^19 = -60x^2 (1 + 1/2x)^19 --> Notice that each term from the (1 + 1/2x)^19 binomial will be multiplied by -60x^2 So we need the x term of the binomial (1 + 1/2)^19 to get the x^3 term, the x term is 19C1 (1/2x) (1)^18 = 19/2 x

Multiply the x^2 part with the x term --> -60x^2 * 19/2x = -570x^3

x^3 straight away Cannot get x^3 from the x^2 part (raising x^2 to any integer power will not give x^3) , so we need to get x^3 from the (1 +1/2x) part: Term needed is: 20 C 0 (-3x^2)^0 (1 + 1/2x)^20 = (1 + 1/2x)^20 Now we just need the coefficient of the x^3 term of this binomial: 20 C 3 (1/2x)^3 (1)^17 = 1140 * 1/8 X^3 = 142.5x^3

Adding the two x^3 terms gives -570x^3 + 142.5x^3 = -427.5x^3