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.