Question

The coefficients of ${x^7}$ in the expansion of $(1 - {x^4}){(1 + x)^9}$ is equal to
A.27
B.-24
C.48
D.-48

Answer 1

Hint : We solve this using binomial expansion. We know in binomial expansion of ${(a + b)^n}$ , the ${(r + 1)^{th}}$ term denoted by ${t_{r + 1}}$ and its given by ${t_{r + 1}}{ = ^n}{C_r}.{a^{n - r}}.{b^r}$ . We convert the above expansion term in the form ${(a + b)^n}$ , then we apply the ${t_{r + 1}}$ formula to find the required coefficient. We know $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ .

Complete step-by-step answer :
We know the binomial expansion ${(a + b)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}}$ , where $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ .
We have, $(1 - {x^4}){(1 + x)^9}$
Expanding we have,
$\Rightarrow 1 \times {(1 + x)^9} - {x^4}{(1 + x)^9}{\text{ - - - - - - (1)}}$
We find the coefficient ${x^7}$ in $1 \times {(1 + x)^9}$ and $- {x^4}{(1 + x)^9}$ , then we add the both.
Now in ${(1 + x)^9}$ we will have ${x^7}$ when we expand,
We know in binomial expansion of ${(a + b)^n}$ , the ${(r + 1)^{th}}$ term denoted by ${t_{r + 1}}$ and its given by $\Rightarrow {t_{r + 1}}{ = ^n}{C_r}.{a^{n - r}}.{b^r}$
${(1 + x)^9}$ we have $n = 9$ , $a = 1$ and $b = x$
$\Rightarrow {t_{r + 1}}{ = ^9}{C_r}.{(1)^{9 - r}}.{(x)^r}$
We will have the coefficient of ${x^7}$ when $r = 7$ .
$\Rightarrow {t_{7 + 1}}{ = ^9}{C_7}.{(1)^{9 - 7}}.{(x)^7}$
$\Rightarrow {t_8}{ = ^9}{C_7}.{x^7}$
The coefficient is $^9{C_7} = \dfrac{{9!}}{{7!(9 - 7)!}}$ (From the formula)
$= \dfrac{{9!}}{{7!(2)!}}$
$= \dfrac{{9 \times 8 \times 7!}}{{7!(2)!}}$
Cancelling $7!$ we have,
$= \dfrac{{9 \times 8}}{2}$
$= 9 \times 4$
$= 36{\text{ - - - - - - (2)}}$
Now taking ${x^4} \times {(1 + x)^9}$ . Following the same procedure as above.
Now in ${x^4} \times {(1 + x)^9}$ we will have ${x^7}$ when we expand,
We know in binomial expansion of ${(a + b)^n}$ , the ${(r + 1)^{th}}$ term denoted by ${t_{r + 1}}$ and its given by $\Rightarrow {t_{r + 1}}{ = ^n}{C_r}.{a^{n - r}}.{b^r}$
${(1 + x)^9}$ we have $n = 9$ , $a = 1$ and $b = x$
$\Rightarrow {t_{r + 1}}{ = ^9}{C_r}.{(1)^{9 - r}}.{(x)^r}$ .
But we have ${x^4}$ multiplied to it.
$\Rightarrow {x^4} \times {t_{r + 1}} = {x^4}{ \times ^9}{C_r}.{(1)^{9 - r}}.{(x)^r}$
$\Rightarrow {x^4}.{t_{r + 1}}{ = ^9}{C_r}{.1^{9 - r}}.{x^{r + 4}}$
We will have the coefficient of ${x^7}$ when $r = 3$ .
$\Rightarrow {t_{3 + 1}}{ = ^9}{C_3}.{(1)^{9 - 3}}.{(x)^7}$
$\Rightarrow {t_4}{ = ^9}{C_3}.{x^7}$
The coefficient is $^9{C_3} = \dfrac{{9!}}{{3!(9 - 3)!}}$ (From the formula)
$= \dfrac{{9!}}{{3!.6!}}$
$= \dfrac{{9 \times 8 \times 7 \times 6!}}{{3!.6!}}$
Cancelling $6!$ we get,
$= \dfrac{{9 \times 8 \times 7}}{{3 \times 2}}$
$= 3 \times 4 \times 7$
$= 84{\text{ - - - - - (3)}}$
Thus we found the coefficient in each term. Now substituting in the equation 1 we have
Coefficient of ${x^7}$ in the expansion of $(1 - {x^4}){(1 + x)^9}$ $= 36 - 84$
$= - 48$
So, the correct answer is “Option D”.

Note : In the expansion ${(a + b)^n}$ to find any coefficient of any term we have a formula that ${t_{r + 1}}{ = ^n}{C_r}.{a^{n - r}}.{b^r}$ . In the above problem we can also find a particular term position in the expansion. That is in the first term we have ${t_8} = 36.{x^7}$ , we can see that it is the eighth term in that expansion. Similarly for the second term the position of ${x^7}$ is fourth. Follow the same procedure for this kind of problem.