Question

The sum of the coefficients of the first 3 terms in the expansion of ${\left( {x - \dfrac{3}{{{x^2}}}} \right)^m},x \ne 0,m \in N$ is 559. Find the terms of the expansion containing ${x^3}$.

Answer 1

From the expansion of ${\left( {a + b} \right)^n}$we get the general term to be ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$and from the given expansion we get the general term to be ${T_{r + 1}} = {}^m{C_r}{x^{m - r}}{\left( { - \dfrac{3}{{{x^2}}}} \right)^r}$and using this we can find the coefficients of the first three terms and equating their sum to 559 we get a quadratic equation and using the quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$we get the value of m and using that in the general formula and equating the power of x to 3 we get the value of r . Hence using these we get the term containing ${x^3}$

Complete step-by-step answer:
We are given that the sum of the coefficients of the first 3 terms in the expansion of ${\left( {x - \dfrac{3}{{{x^2}}}} \right)^m},x \ne 0,m \in N$ is 559
From the expansion of ${\left( {a + b} \right)^n}$ we get the general term to be
$\Rightarrow {T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$
Using this we get the general term of the expansion ${\left( {x - \dfrac{3}{{{x^2}}}} \right)^m},x \ne 0,m \in N$
Here $a = x,b = \dfrac{{ - 3}}{{{x^2}}},n = m$
Therefore the general term is
$\Rightarrow {T_{r + 1}} = {}^m{C_r}{x^{m - r}}{\left( { - \dfrac{3}{{{x^2}}}} \right)^r}$
Now the first three terms are ${T_1},{T_2},{T_3}$
So let's find the values of the first three terms using the general term formula
$\Rightarrow {T_1} = {}^m{C_0}{x^{m - 0}}{\left( { - \dfrac{3}{{{x^2}}}} \right)^0} \\\ \Rightarrow {T_1} = {x^m} \\\$
$\Rightarrow {T_2} = {}^m{C_1}{x^{m - 1}}{\left( { - \dfrac{3}{{{x^2}}}} \right)^1} \\\ \Rightarrow {T_2} = m{x^{m - 1}}\left( { - \dfrac{3}{{{x^2}}}} \right) \\\ \Rightarrow {T_2} = - 3m{x^{m - 3}} \\\$
$\Rightarrow {T_3} = {}^m{C_2}{x^{m - 2}}{\left( { - \dfrac{3}{{{x^2}}}} \right)^2} \\\ \Rightarrow {T_3} = \dfrac{{m(m - 1)}}{2}{x^{m - 2}}\left( { - \dfrac{{{3^2}}}{{{x^4}}}} \right) \\\ \Rightarrow {T_3} = - {3^2}*\dfrac{{m(m - 1)}}{2}{x^{m - 6}} \\\$
Since we are given that the sum of the coefficients of the first three terms are 559
$\Rightarrow 1 + \left( { - 3m} \right) + \left( {{3^2}*\dfrac{{m\left( {m - 1} \right)}}{2}} \right) = 559 \\\ \Rightarrow 1 - 3m + \left( {\dfrac{{9m\left( {m - 1} \right)}}{2}} \right) = 559 \\\ \Rightarrow 1 - 3m + \left( {\dfrac{{9{m^2} - 9m}}{2}} \right) = 559 \\\ \Rightarrow \dfrac{{2 - 6m + 9{m^2} - 9m}}{2} = 559 \\\ \Rightarrow 2 - 15m + 9{m^2} = 1118 \\\ \Rightarrow 9{m^2} - 15m + 2 - 1118 = 0 \\\ \Rightarrow 9{m^2} - 15m - 1116 = 0 \\\$
We can use the quadratic formula to find the way of m

$$\Rightarrow \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\\ \Rightarrow m = \dfrac{{15 \pm \sqrt {{{15}^2} - 4\left( 9 \right)\left( { - 1116} \right)} }}{{2\left( 9 \right)}} \\\ \Rightarrow m = \dfrac{{15 \pm \sqrt {225 + 40176} }}{{18}} \\\ \Rightarrow m = \dfrac{{15 \pm \sqrt {40401} }}{{18}} = \dfrac{{15 \pm 201}}{{18}} \\\ \Rightarrow m = \dfrac{{3\left( {5 \pm 67} \right)}}{{18}} = \dfrac{{\left( {5 \pm 67} \right)}}{6} \\\ \Rightarrow m = \dfrac{{5 + 67}}{6},\dfrac{{5 - 67}}{6} \\\ \Rightarrow m = \dfrac{{72}}{6},\dfrac{{ - 62}}{6} \\\ \Rightarrow m = 12,\dfrac{{ - 31}}{3} \\\$$

Since we are given m belongs to natural numbers m = 12 is accepted
Since we need the term containing ${x^3}$

$$\Rightarrow {T_{r + 1}} = {}^{12}{C_r}{x^{12 - r}}{\left( { - \dfrac{3}{{{x^2}}}} \right)^r} \\\ \Rightarrow {T_{r + 1}} = {}^{12}{C_r}{x^{12 - r - 2r}}{\left( { - 3} \right)^r} \\\ \Rightarrow {T_{r + 1}} = {}^{12}{C_r}{x^{12 - 3r}}{\left( { - 3} \right)^r} \\\$$

Since we need to find the term containing ${x^3}$ let's find the value of r first
$\Rightarrow {x^{12 - 3r}} = {x^3} \\\ \Rightarrow 12 - 3r = 3 \\\ \Rightarrow 12 - 3 = 3r \\\ \Rightarrow 9 = 3r \\\ \Rightarrow r = \dfrac{9}{3} = 3 \\\$
Now using this in the general formula we get

$$\Rightarrow {T_{3 + 1}} = {}^{12}{C_3}{x^{12 - 3}}{\left( { - \dfrac{3}{{{x^2}}}} \right)^3} \\\ \Rightarrow {T_4} = \dfrac{{12*11*10}}{{1*2*3}}{x^9}\left( { - \dfrac{{{3^3}}}{{{x^6}}}} \right) \\\ \Rightarrow {T_4} = 2*11*10*{x^3}\left( { - 27} \right) \\\ \Rightarrow {T_4} = - 5940{x^3} \\\$$

Hence we obtained the term containing ${x^3}$.

Note: There are n+1 terms in the expansion of ${(x + y)^n}$
The degree of each term is n
The powers on x begin with n and decrease to 0
The powers on y begin with 0 and increase to n
The coefficients are symmetric