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The expansion of (1+px)^q in ascending powers of x begins 1 - 6x + 24x² + ..., where p and q are constants. (a) Find the values of p and q. (b) State the set of values of x for which the expansion is valid. (c) Find the coefficient of x³ in the expansion.

Mathematics

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Answers

3 parts
a

p=2,q=−3p=2, q=-3

b

−12<x<12-\frac{1}{2} < x < \frac{1}{2}

c
-80

Step-by-step solution

7 steps

Part a· 3 steps

To find the values of pp and qq, we use the binomial expansion for (1+ax)n(1+ax)^n, which is given by

(1+ax)n=1+n(ax)+n(n−1)2!(ax)2+n(n−1)(n−2)3!(ax)3+…(1+ax)^n = 1 + n(ax) + \frac{n(n-1)}{2!}(ax)^2 + \frac{n(n-1)(n-2)}{3!}(ax)^3 + \dots

In this problem, we have (1+px)q(1+px)^q. So, a=pa=p and n=qn=q.
The expansion is:

(1+px)q=1+q(px)+q(q−1)2(px)2+…(1+px)^q = 1 + q(px) + \frac{q(q-1)}{2}(px)^2 + \dots (1+px)q=1+qpx+q(q−1)2p2x2+…(1+px)^q = 1 + qpx + \frac{q(q-1)}{2}p^2x^2 + \dots

We are given that the expansion begins 1−6x+24x2+…1 - 6x + 24x^2 + \dots.
Comparing the coefficients of xx:

  1. Equate the coefficient of xx.

    qp=−6(1)qp = -6 \quad (1)

  2. Equate the coefficient of x2x^2.

    q(q−1)2p2=24(2)\frac{q(q-1)}{2}p^2 = 24 \quad (2)

  3. Solve the system of equations.

    From equation (1), we can express pp in terms of qq:

    p=−6qp = -\frac{6}{q}

    Substitute this into equation (2):

    q(q−1)2(−6q)2=24\frac{q(q-1)}{2}\left(-\frac{6}{q}\right)^2 = 24 q(q−1)2(36q2)=24\frac{q(q-1)}{2}\left(\frac{36}{q^2}\right) = 24 18q(q−1)q2=24\frac{18q(q-1)}{q^2} = 24 18(q−1)q=24\frac{18(q-1)}{q} = 24 18(q−1)=24q18(q-1) = 24q 18q−18=24q18q - 18 = 24q −18=6q-18 = 6q q=−186q = -\frac{18}{6} q=−3q = -3

    Now substitute q=−3q=-3 back into p=−6qp = -\frac{6}{q}:

    p=−6−3p = -\frac{6}{-3} p=2p = 2

    The values are:

    p=2,q=−3\boxed{p=2, q=-3}

Part b

−12<x<12-\frac{1}{2} < x < \frac{1}{2}

The binomial expansion (1+Ax)n(1+Ax)^n is valid for ∣Ax∣<1|Ax| < 1.

In this case, we have (1+px)q(1+px)^q, and we found p=2p=2.

  1. Apply the condition for validity.

    ∣px∣<1|px| < 1 ∣2x∣<1|2x| < 1

  2. Solve the inequality for xx.

    −1<2x<1-1 < 2x < 1 −12<x<12-\frac{1}{2} < x < \frac{1}{2}

    The set of values for which the expansion is valid is:

    −12<x<12\boxed{-\frac{1}{2} < x < \frac{1}{2}}

Part c-80

To find the coefficient of x3x^3, we use the general term for x3x^3

q(q−1)(q−2)3!(px)3\frac{q(q-1)(q-2)}{3!}(px)^3

The coefficient of x3x^3 is q(q−1)(q−2)6p3\frac{q(q-1)(q-2)}{6}p^3.

  1. Substitute the values of p=2p=2 and q=−3q=-3.

    Coefficient of x3=(−3)(−3−1)(−3−2)6(2)3\text{Coefficient of } x^3 = \frac{(-3)(-3-1)(-3-2)}{6}(2)^3

  2. Simplify the expression.

    =(−3)(−4)(−5)6(8)= \frac{(-3)(-4)(-5)}{6}(8) =−606(8)= \frac{-60}{6}(8) =−10(8)= -10(8) =−80= -80

    The coefficient of x3x^3 is:

    −80\boxed{-80}
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