Question

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

Answer 1

We will first expand the expression ${\left( {x - \dfrac{3}{{{x^2}}}} \right)^m},x \ne 0$ using binomial expansion property and then put the sum of first three terms equal to 559 as given in the question and simplify the expression. After converting the expression into equation, we will find the values of $m$ by using middle term splitting method. Then we will find the term using ${T_{r + 1}}$ expression by solving for $r$ and as we need to find the term containing ${x^3}$ so, we will put the value of $r$ in ${T_{r + 1}}$ and get the desired result.

Complete step by step Answer:

We will first consider the given expression, ${\left( {x - \dfrac{3}{{{x^2}}}} \right)^m},x \ne 0$.
Now, we will expand the expression using binomial expansion,
Thus, we get,

$$\Rightarrow {\left( {x - \dfrac{3}{{{x^2}}}} \right)^m} = {}^m{C_0}{\left( x \right)^m}{\left( {\dfrac{{ - 3}}{{{x^2}}}} \right)^0} + {}^m{C_1}{\left( x \right)^{m - 1}}{\left( {\dfrac{{ - 3}}{{{x^2}}}} \right)^1} + {}^m{C_2}{\left( x \right)^{m - 2}}{\left( {\dfrac{{ - 3}}{{{x^2}}}} \right)^2} + .... \\\ \Rightarrow {\left( {x - \dfrac{3}{{{x^2}}}} \right)^m} = {}^m{C_0}{\left( x \right)^m} + \left( { - 3} \right){}^m{C_1}{\left( x \right)^{m - 1 - 2}} + 9{}^m{C_2}{\left( x \right)^{m - 2 - 4}} + ..... \\\$$

Now, as given in the question that the sum of three terms is equal to 559 so, we will put the sum of the first three terms equal to 559.
Thus, we get,

$$\Rightarrow {}^m{C_0} + \left( { - 3} \right){}^m{C_1} + 9{}^m{C_2} = 559 \\\ \Rightarrow 1 - 3m + 9\dfrac{{m\left( {m - 1} \right)}}{2} = 559 \\\ \Rightarrow 2 - 6m + 9{m^2} - 9m = 2\left( {559} \right) \\\ \Rightarrow 9{m^2} - 15m + 2 - 1118 = 0 \\\ \Rightarrow 9{m^2} - 15m - 1116 = 0 \\\$$

Now, we will solve the obtained equation using the middle term splitting method,

$$\Rightarrow 3{m^2} - 15m - 372 = 0 \\\ \Rightarrow 3{m^2} - 36m + 31m - 372 = 0 \\\ \Rightarrow 3m\left( {m - 12} \right) + 31\left( {m - 12} \right) = 0 \\\ \Rightarrow \left( {m - 12} \right)\left( {3m + 31} \right) = 0 \\\$$

Now, we will apply the zero-factor property to find the values of $m$,
$\Rightarrow m - 12 = 0$ and $3m + 31 = 0$
$\Rightarrow m = 12$ and $m = \dfrac{{ - 31}}{3}$
Here, we will not consider the negative value of $m$ so, we will ignore it and choose $m = 12$.
Thus, we will substitute the value of $m$ in the given expression and we get, ${\left( {x - \dfrac{3}{{{x^2}}}} \right)^{12}},x \ne 0$
Next, we will use the expression to find the terms using the formula, ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$, that is ${T_{r + 1}} = {}^{12}{C_r}{x^{12 - r}}{\left( {\dfrac{{ - 3}}{{{x^2}}}} \right)^r}$ and simplify it.
Thus, we get,

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

Now, as we need to find the term containing ${x^3}$, so, we will put the power of $x$ equal to 0.
Thus, we get,

$$\Rightarrow 12 - 3r = 0 \\\ \Rightarrow r = 3 \\\$$

Now, we will substitute the value of $r$ in the expression ${T_{r + 1}} = {\left( { - 1} \right)^r}{\left( 3 \right)^r}{}^{12}{C_r}{x^{12 - 3r}}$ to find the term of expansion of ${x^3}$.

$$\Rightarrow {T_4} = {T_{3 + 1}} \\\ \Rightarrow {T_4} = {\left( { - 1} \right)^3}{3^3}{}^{12}{C_3}{x^3} \\\ \Rightarrow {T_4} = - 27{}^{12}{C_3}{x^3} \\\ \Rightarrow {T_4} = - 5940{x^3} \\\$$

Hence, we can conclude that the term of expansion containing ${x^3}$ is $- 5940{x^3}$.

Note: We need to remember the formula of expanding the expression using binomial expansion. We have evaluated the value of $m$ using the fact that the sum of the first three terms is 559. We have to remember that to calculate the terms we have to use the form ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{\left( b \right)^r}$. As we have to find the term of expansion containing ${x^3}$ so, we need to put the power on $x$ equal to 0.