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.