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Obtain the first four terms of the binomial expansion of (1 - 3/2x)⁻² and find the set of values of x for which the expansion is valid. Hence, deduce the first four terms of the binomial expansion of

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3 parts
1

1+3x+274x2+272x31 + 3x + \frac{27}{4}x^2 + \frac{27}{2}x^3

2

−23<x<23-\frac{2}{3} < x < \frac{2}{3}

3

14+34x+2716x2+278x3\frac{1}{4} + \frac{3}{4}x + \frac{27}{16}x^2 + \frac{27}{8}x^3

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Here is the solution for Question 4 (ii):

Part 4

(ii)

Obtain the first four terms of the binomial expansion of (1−32x)−2\left(1 - \frac{3}{2}x\right)^{-2} and find the set of values of xx for which the expansion is valid. Hence, deduce the first four terms of the binomial expansion of (2−3x)−2(2 - 3x)^{-2}.

Part 1

1+3x+274x2+272x31 + 3x + \frac{27}{4}x^2 + \frac{27}{2}x^3

Obtain the first four terms of the binomial expansion of (1−32x)−2\left(1 - \frac{3}{2}x\right)^{-2}.

The binomial expansion for (1+u)n(1+u)^n is given by:

(1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+…(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \dots

In this case, u=−32xu = -\frac{3}{2}x and n=−2n = -2.

  1. Calculate the first term.

    The first term is 1.

  2. Calculate the second term.

    nu=(−2)(−32x)=3xnu = (-2)\left(-\frac{3}{2}x\right) = 3x

  3. Calculate the third term.

    n(n−1)2!u2=(−2)(−2−1)2×1(−32x)2\frac{n(n-1)}{2!}u^2 = \frac{(-2)(-2-1)}{2 \times 1}\left(-\frac{3}{2}x\right)^2 =(−2)(−3)2(94x2)= \frac{(-2)(-3)}{2}\left(\frac{9}{4}x^2\right) =3(94x2)=274x2= 3\left(\frac{9}{4}x^2\right) = \frac{27}{4}x^2

  4. Calculate the fourth term.

    n(n−1)(n−2)3!u3=(−2)(−2−1)(−2−2)3×2×1(−32x)3\frac{n(n-1)(n-2)}{3!}u^3 = \frac{(-2)(-2-1)(-2-2)}{3 \times 2 \times 1}\left(-\frac{3}{2}x\right)^3 =(−2)(−3)(−4)6(−278x3)= \frac{(-2)(-3)(-4)}{6}\left(-\frac{27}{8}x^3\right) =−246(−278x3)= \frac{-24}{6}\left(-\frac{27}{8}x^3\right) =−4(−278x3)=272x3= -4\left(-\frac{27}{8}x^3\right) = \frac{27}{2}x^3

    The first four terms of the expansion of (1−32x)−2\left(1 - \frac{3}{2}x\right)^{-2} are 1+3x+274x2+272x3\boxed{1 + 3x + \frac{27}{4}x^2 + \frac{27}{2}x^3}.

Part 2

−23<x<23-\frac{2}{3} < x < \frac{2}{3}

Find the set of values of xx for which the expansion is valid.

The binomial expansion (1+u)n(1+u)^n is valid when ∣u∣<1|u| < 1.
Here, u=−32xu = -\frac{3}{2}x.
So, we must have:

∣−32x∣<1\left|-\frac{3}{2}x\right| < 1 32∣x∣<1\frac{3}{2}|x| < 1 ∣x∣<23|x| < \frac{2}{3}

The set of values of xx for which the expansion is valid is −23<x<23\boxed{-\frac{2}{3} < x < \frac{2}{3}}.

Part 3

14+34x+2716x2+278x3\frac{1}{4} + \frac{3}{4}x + \frac{27}{16}x^2 + \frac{27}{8}x^3

Deduce the first four terms of the binomial expansion of (2−3x)−2(2 - 3x)^{-2}.

  1. Rewrite (2−3x)−2(2 - 3x)^{-2} in the form A(1+u)nA(1+u)^n.

    (2−3x)−2=[2(1−32x)]−2(2 - 3x)^{-2} = \left[2\left(1 - \frac{3}{2}x\right)\right]^{-2} =2−2(1−32x)−2= 2^{-2}\left(1 - \frac{3}{2}x\right)^{-2} =14(1−32x)−2= \frac{1}{4}\left(1 - \frac{3}{2}x\right)^{-2}

  2. Substitute the expansion from Part 1 into this expression.

    =14(1+3x+274x2+272x3+… )= \frac{1}{4}\left(1 + 3x + \frac{27}{4}x^2 + \frac{27}{2}x^3 + \dots\right)

  3. Multiply each term by 14\frac{1}{4}.

    =14(1)+14(3x)+14(274x2)+14(272x3)+…= \frac{1}{4}(1) + \frac{1}{4}(3x) + \frac{1}{4}\left(\frac{27}{4}x^2\right) + \frac{1}{4}\left(\frac{27}{2}x^3\right) + \dots =14+34x+2716x2+278x3+…= \frac{1}{4} + \frac{3}{4}x + \frac{27}{16}x^2 + \frac{27}{8}x^3 + \dots

    The first four terms of the binomial expansion of (2−3x)−2(2 - 3x)^{-2} are 14+34x+2716x2+278x3\boxed{\frac{1}{4} + \frac{3}{4}x + \frac{27}{16}x^2 + \frac{27}{8}x^3}.

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