Question

Find the coefficient of \[{x^{13}}\] in the expansion of \[{(1 - x)^5}.x.{(1 + x + {x^2} + {x^3})^4}\] .

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 it's 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, \[{(1 - x)^5} \times x \times {(1 + x + {x^2} + {x^3})^4}\] ------ (1)
As we can see, in the second term \[1 + x + {x^2} + {x^3}\] are in G.P.
Sum of n terms \[{S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}\] .
In \[{(1 + x + {x^2} + {x^3})^4}\] we have \[a = 1\] , \[r = x\] and \[n = 4\] .
 \[ \Rightarrow {S_4} = {\left( {\dfrac{{1 - {x^4}}}{{1 - x}}} \right)^4}\]
 \[ \Rightarrow {S_4} = \dfrac{{{{(1 - {x^4})}^4}}}{{{{(1 - x)}^4}}}\]
Substituting in equation (1), we get
 \[{(1 - x)^5} \times x \times {(1 + x + {x^2} + {x^3})^4}\]
 \[ \Rightarrow = {(1 - x)^5} \times x \times \dfrac{{{{(1 - {x^4})}^4}}}{{{{(1 - x)}^4}}}\]
Cancelling terms
 \[ \Rightarrow {(1 - x)^1} \times x \times {(1 - {x^4})^4}\] .
In \[{(1 - x)^1}\] in this expansion we only have \[x\] . Meaning that we cannot have \[{x^{13}}\] term in this expansion. So we neglect \[{(1 - x)^1}\] term. Now in \[x \times {(1 - {x^4})^4}\] , in the expansion of \[{(1 - {x^4})^4}\] we get \[{x^4},{x^8},{x^{12}}...\] and we have \[x\] multiplying for each of the term \[{x^4},{x^8},{x^{12}}...\] we will get \[{x^{13}}\] when \[{x^{12}}\] multiplied by \[x\] .
Hence, we take only the terms \[x \times {(1 - {x^4})^4}\] ------ (2)
We know in binomial expansion of \[{(a + b)^n}\] , the \[{(r + 1)^{th}}\] term denoted by \[{t_{r + 1}}\] and its given by \[{t_{r + 1}}{ = ^n}{C_r}.{a^{n - r}}.{b^r}\]
 \[{(1 - {x^4})^4}\] we have \[n = 4\] , \[a = 1\] and \[b = - {x^4}\]
 \[{t_{r + 1}}{ = ^4}{C_r}.{(1)^{4 - r}}.{( - {x^4})^r}\]
Substituting in equation (2).
 \[ \Rightarrow x \times {{\text{ }}^4}{C_r}.{(1)^{4 - r}}.{( - {x^4})^r}\]
 \[{ \Rightarrow ^4}{C_r}.{(1)^{4 - r}}.{( - {x^4})^r} \times {x^1}\]
We need to put the \[r\] value by guessing, so that we need to get the term \[{x^{13}}\] .
If we put \[r = 3\] we get the term \[{x^{13}}\] .
 \[{ \Rightarrow ^4}{C_3}.{(1)^{4 - 3}}.{( - {x^4})^3} \times {x^1}\]
 \[ \Rightarrow { - ^4}{C_3}.{(1)^{4 - 3}}.{\text{ }}{x^{12}}.{x^1}\]
 \[ \Rightarrow { - ^4}{C_3}.{\text{ }}{x^{13}}\] ---- (3)
Now, \[ \Rightarrow {(^4}{C_3}) = \dfrac{{4!}}{{3!(4 - 3)!}}\]
 \[ \Rightarrow {(^4}{C_3}) = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1(1)!}}\]
 \[ \Rightarrow {(^4}{C_3}) = 4\]
Now substituting in equation (3).
 \[ \Rightarrow - 4.{\text{ }}{x^{13}}\]
Hence the coefficient of \[{x^{13}}\] is -4.
So, the correct answer is “-4”.

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 \[{t_{r + 1}} = {t_4}\] because \[r = 3\] . That is \[ - 4.{x^{13}}\] is fourth term in the expansion of \[{(1 - x)^5}.x.{(1 + x + {x^2} + {x^3})^4}\] . Follow the same procedure as above for this kind of problem.