Question

How to expand ${(3x - 2)^{ - 2}}$ in ascending powers of $\dfrac{1}{x}$ , stating the first four non-zero terms and the value of $x$ for which the expansion is valid?

Answer 1

Hint : Use of Binomial theorem to expand the given equation. First take out $3$ common from the equation then expand the values as need using the basic formula of Binomial theorem that is ${(1 + x)^n} = 1 + nx + \dfrac{{n(n - 1){x^2}}}{{2!}} + \dfrac{{n(n - 1)(n - 2){x^3}}}{{3!}} + ..$ just put the values accordingly and get the results.

Complete step-by-step answer :
We are given the equation ${\left( {3 - x} \right)^{ - 2}}$ to expand in ascending orders of $\dfrac{1}{x}$ , with four non-zero terms.
So, we are going to use the binomial theorem and from Binomial theorem we know that: ${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right){x^3}}}{{3!}} + ..$ (i)
Since we want only four non zero terms so we can stop at fourth term only.

Case $1$ :
Since we need $\left( {1 + x} \right)$ so take $3$ common from the given equation and we get:
${3^{ - 2}}{\left( {1 - \dfrac{x}{3}} \right)^{ - 2}}$ .
On further solving we get:
$\dfrac{1}{{{3^2}}}{\left( {1 - \dfrac{x}{3}} \right)^{ - 2}} = \dfrac{1}{9}{\left( {1 - \dfrac{x}{3}} \right)^{ - 2}}$
On equating the above term and equation (i) we get:
$x = \dfrac{{ - x}}{3}$ and $n = - 2$
Just put these values in the Expansion formula (i) and we get:
${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right){x^3}}}{{3!}} + .. \\\ {\left( {1 + \dfrac{{ - x}}{3}} \right)^{ - 2}} = 1 + \left( { - 2} \right)\left( {\dfrac{{ - x}}{3}} \right) + \dfrac{{\left( { - 2} \right)\left( { - 2 - 1} \right){{\left( {\dfrac{{ - x}}{3}} \right)}^2}}}{{2!}} + \dfrac{{\left( { - 2} \right)\left( { - 2 - 1} \right)\left( { - 2 - 2} \right){{\left( {\dfrac{{ - x}}{3}} \right)}^3}}}{{3!}} \;$
Simplifying it, and we get:
${\left( {1 + \dfrac{{ - x}}{3}} \right)^{ - 2}} = \left( {1 + \dfrac{{2x}}{3} + \dfrac{{{x^2}}}{3} + \dfrac{{4{x^3}}}{{27}}} \right)$
Substituting this value in $\dfrac{1}{9}{\left( {1 + \dfrac{{ - x}}{3}} \right)^{ - 2}}$ :
$\dfrac{1}{9}{\left( {1 + \dfrac{{ - x}}{3}} \right)^{ - 2}} = \dfrac{1}{9}\left( {1 + \dfrac{{2x}}{3} + \dfrac{{{x^2}}}{3} + \dfrac{{4{x^3}}}{{27}}} \right)$
Opening the parenthesis on the right side and we get:
$\dfrac{1}{9}{\left( {1 + \dfrac{{ - x}}{3}} \right)^{ - 2}} = \dfrac{1}{9} + \dfrac{{2x}}{{27}} + \dfrac{{{x^2}}}{{27}} + \dfrac{{4{x^3}}}{{243}}$ .
Since the value of $x$ is given by $\dfrac{{ - x}}{3}$ .And $\left| {\dfrac{{ - x}}{3}} \right| < 1$ So the value of $x$ for which the expansion is valid is $\left| x \right| < 3$ .

Case $2$ :
Next, we repeat the same steps with some changes:
Since we need $\left( {1 + x} \right)$ so take $- x$ common now from the given equation and we get:
${\left( { - x} \right)^{ - 2}}{\left( {1 - \dfrac{3}{x}} \right)^{ - 2}}$
On further solving we get:
$\dfrac{1}{{{x^2}}}{\left( {1 - \dfrac{3}{x}} \right)^{ - 2}} = \dfrac{1}{{{x^2}}}{\left( {1 - \dfrac{3}{x}} \right)^{ - 2}}$
On equating the above term and equation (i) we get:
$x = \dfrac{{ - 3}}{x}$ and $n = - 2$
Just put these values in the Expansion formula (i) and we get:
${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right){x^3}}}{{3!}} + .. \\\ {\left( {1 + \dfrac{{ - 3}}{x}} \right)^{ - 2}} = 1 + \left( { - 2} \right)\left( {\dfrac{{ - 3}}{x}} \right) + \dfrac{{\left( { - 2} \right)\left( { - 2 - 1} \right){{\left( {\dfrac{{ - 3}}{x}} \right)}^2}}}{{2!}} + \dfrac{{\left( { - 2} \right)\left( { - 2 - 1} \right)\left( { - 2 - 2} \right){{\left( {\dfrac{{ - 3}}{x}} \right)}^3}}}{{3!}} \\\$
Simplifying it, and we get:
${\left( {1 + \dfrac{{ - 3}}{x}} \right)^{ - 2}} = \left( {1 + \dfrac{6}{x} + \dfrac{{27}}{{{x^2}}} + \dfrac{{108}}{{{x^3}}}} \right)$
Substituting this value in $\dfrac{1}{{{x^2}}}{\left( {1 - \dfrac{3}{x}} \right)^{ - 2}}$ :
$\dfrac{1}{{{x^2}}}{\left( {1 + \dfrac{{ - 3}}{x}} \right)^{ - 2}} = \dfrac{1}{{{x^2}}}\left( {1 + \dfrac{6}{x} + \dfrac{{27}}{{{x^2}}} + \dfrac{{108}}{{{x^3}}}} \right)$
Opening the parenthesis on the right, side and we get:
$\dfrac{1}{{{x^2}}}{\left( {1 + \dfrac{{ - 3}}{x}} \right)^{ - 2}} = \dfrac{1}{{{x^2}}} + \dfrac{6}{{{x^3}}} + \dfrac{{27}}{{{x^4}}} + \dfrac{{108}}{{{x^5}}}$
Since the value of $x$ is given by $\dfrac{{ - 3}}{x}$.And $\left| {\dfrac{{ - 3}}{x}} \right| < 1$ . So the value of $x$ for which the expansion is valid is $\left| x \right| > 3$ .

Note : There can be an error in putting the values in the formula and get the results.
Don’t forget to multiply the coefficients in the expanded formula.
Always check up to which term values are needed.
Also remember that we are supposed to take minus power out along with common terms.