Question

The coefficient of ${x^7}$ in the expansion of ${\left( {1 - x - {x^2} + {x^3}} \right)^6}$ is

$${\text{(A) }}144 \\\ {\text{(B) }} - 132 \\\ {\text{(C) }} - 144 \\\ {\text{(D) }}132 \\\$$

Answer 1

This is a problem related to Binomial Theorem. To solve it, we shall have to factorise in simple terms so that Binomial Theorem can be applied. The general expression for Binomial Theorem is ${(a + b)^n} = \sum\limits_{k = 0}^n {{}^n{C_k}{a^{n - k}}} {b^k}$, ${(a - b)^n} = \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{a^{n - k}}} {b^k}$.Using these theorems we try to get the anwer.

Complete step-by-step answer:
First, write the expression given in the question as below,
${\left( {1 - x - {x^2} + {x^3}} \right)^6}$
This equation can be re-written by factorising in terms of $(1 - x)$ we get the following expression.
${\left( {(1 - x) - {x^2}(1 - x)} \right)^6} \\\ = {(1 - x)^6}{(1 - {x^2})^6}{\text{ }}...............{\text{ (1)}}$
After factorising, we have got two expressions. These expressions can be expanded with Binomial Expansion ${(a - b)^n} = \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{a^{n - k}}} {b^k}$.
In the first expression of eq. (1) ,$a = 1,{\text{ and }}b = x$ and $k = p,{\text{ ranging from 0 to 6, thus }}n = 6$here, so the expansion of first expression of the eq. (1) will be
{(1 - x)^6} = \left\\{ {\sum\limits_{p = 0}^6 {{{( - 1)}^p}{}^6{C_p}.{x^p}} } \right\\} ……………… (2)
In the similar way, the second expression of eq. (1) can be written where $a = 1,{\text{ and }}b = {x^2}$ and $k = q,{\text{ ranging from 0 to 6, thus }}n = 6$here, so the expansion of second expression of the eq. (1) will be
{(1 - {x^2})^6} = \left\\{ {\sum\limits_{q = 0}^6 {{{( - 1)}^q}{}^6{C_q}.{x^{2q}}} } \right\\}{\text{ }}...........{\text{ (3)}}
Replacing the values of ${(1 - x)^6}$ and ${(1 - {x^2})^6}$ from eq. (2) and (3) respectively, in the eq. (1), we get the following expression,
= \left\\{ {\sum\limits_{p = 0}^6 {{{( - 1)}^p}{}^6{C_p}.{x^p}} } \right\\}.\left\\{ {\sum\limits_{q = 0}^6 {{{( - 1)}^q}{}^6{C_q}.{x^{2q}}} } \right\\}
This can, again, be simplified by rearranging the terms inside the $\sum {}$as
$= \sum\limits_{p = 0}^6 {\sum\limits_{q = 0}^6 {{{( - 1)}^p}{}^6{C_p}.{x^p}} .{{( - 1)}^q}{}^6{C_q}.{x^{2q}}}$
Adding the powers of $x$ and (-1), we will get the following expression,
$= \sum\limits_{p = 0}^6 {\sum\limits_{q = 0}^6 {{{( - 1)}^{p + q}}{}^6{C_p}.} {}^6{C_q}.{x^{p + 2q}}} {\text{ }}..........{\text{(4)}}$
Now to find out the coefficient of ${x^7}$, we have an equation $p + 2q = 7$.
As this is a linear equation with two variables, it can be solved by putting the values of $p$, we will get the values for $q$. Now, following are the cases that will happen for different values of $p$.
${\text{When }}p = 1, \\\ 1 + 2q = 7 \\\ q = 3 \\\$ ${\text{When }}p = 3, \\\ 3 + 2q = 7 \\\ q = 2 \\\$ ${\text{When }}p = 5, \\\ 5 + 2q = 7 \\\ q = 1$
Now, putting these values of $p$ and $q$ in equation (4), we will get the following expression,
$= {( - 1)^{1 + 3}}{}^6{C_1}.{}^6{C_3}.{x^7} + {( - 1)^{3 + 2}}{}^6{C_3}.{}^6{C_2}.{x^7} + {( - 1)^{5 + 1}}{}^6{C_5}.{}^6{C_1}.{x^7}$
As we know that,${}^n{C_r} = \dfrac{{n!}}{{r! \times (n - r)!}}$ therefore, solving the above expression,
$= {( - 1)^4} \times {}^6{C_1}.{}^6{C_3}.{x^7} + {( - 1)^5} \times {}^6{C_3}.{}^6{C_2}.{x^7} + {( - 1)^6} \times {}^6{C_5}.{}^6{C_1}.{x^7} \\\ = [(\dfrac{{6!}}{{5! \times 1!}} \times \dfrac{{6!}}{{3! \times 3!}}) - (\dfrac{{6!}}{{3! \times 3!}} \times \dfrac{{6!}}{{4! \times 2!}}) + (\dfrac{{6!}}{{5! \times 1!}} \times \dfrac{{6!}}{{1! \times 5!}})]{x^7} \\\$
From the above expression, coefficient of ${x^7}$can be written as
$= [(\dfrac{{6!}}{{5! \times 1!}} \times \dfrac{{6!}}{{3! \times 3!}}) - (\dfrac{{6!}}{{3! \times 3!}} \times \dfrac{{6!}}{{4! \times 2!}}) + (\dfrac{{6!}}{{5! \times 1!}} \times \dfrac{{6!}}{{1! \times 5!}})]$
Solving the above expression, we will get,

= [(\dfrac{{6 \times 5!}}{{5! \times 1}} \times \dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 3 \times 2 \times 1}}) - (\dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 3 \times 2 \times 1}} \times \dfrac{{6 \times 5 \times 4!}}{{4! \times 2 \times 1}}) + (\dfrac{{6 \times 5!}}{{5! \times 1}} \times \dfrac{{6 \times 5!}}{{1 \times 5!}})] \\\ = 6 \times 5 \times 4 - (5 \times 4 \times 5 \times 3) + 6 \times 6 \\\ = 120 - 300 + 36 \\\ = - 144 $$ **So, the correct answer is “Option C”.** **Note:** When the questions are asked for higher powers of variables in the equation given, you should understand that the problem will be solved only by binomial theorem. Binomial expansion is the easiest way to solve them otherwise these problems can be solved by thorough expansion of the function which is very lengthy and time consuming.