Question

Find the coefficient of \[{x^{32}}\]and \[{x^{ - 17}}\]in\[{\left( {{x^4} - \dfrac{1}{{{x^3}}}} \right)^{15}}\].

Answer 1

Hint: General term of expansion of \[{\left( {x + y} \right)^n}\]for the \[{\left( {r + 1} \right)^{th}}\]term is given as \[{T_{r + 1}} = {\left( { - 1} \right)^r}{}^n{C_r}{x^{n - r}}{y^r}\]
Where n is the power to which the binomial is raised and r is the number of the terms which varies from 0, 1, 2, 3,…. , n.
In this question a polynomial is given with the power 15, we will compare this polynomial with the general term of expansion and will find its \[{\left( {r + 1} \right)^{th}}\]and by comparing its power to the given coefficients we will find the terms and then we will find its coefficients.

Complete step-by-step answer:
Given the polynomial is \[{\left( {{x^4} - \dfrac{1}{{{x^3}}}} \right)^{15}}\] where \[n = 15\],
We know the general term of expansion of \[{\left( {x + y} \right)^n}\]for the \[{\left( {r + 1} \right)^{th}}\] term is given as \[{T_{r + 1}} = {\left( { - 1} \right)^r}{}^n{C_r}{x^{n - r}}{y^r}\]
By putting the values \[n = 15\], \[x = {x^4}\]and \[y = \dfrac{1}{{{x^3}}}\] in this general term of expansion we can write
\[{T_{r + 1}} = {\left( { - 1} \right)^r}{}^{15}{C_r}{\left( {{x^4}} \right)^{15 - r}}{\left( {\dfrac{1}{{{x^3}}}} \right)^r} - - (i)\]
Hence by further solving we can write
\[
\Rightarrow {T_{r + 1}} = {\left( { - 1} \right)^r}{}^{15}{C_r}{\left( {{x^4}} \right)^{15 - r}}{\left( {{x^{ - 3}}} \right)^r} \\
   = {\left( { - 1} \right)^r}{}^{15}{C_r}{x^{4\left( {15 - r} \right)}}{x^{ - 3r}} \\
   = {\left( { - 1} \right)^r}{}^{15}{C_r}{x^{60 - 4r - 3r}} \\
   = {\left( { - 1} \right)^r}{}^{15}{C_r}{x^{60 - 7r}} \\
 \]
Now for \[{x^{32}}\]from \[{x^{60 - 7r}}\]we can write
\[60 - 7r = 32\]
Hence by further solving we get
\[
\Rightarrow 60 - 32 = 7r \\
\Rightarrow 7r = 28 \\
\Rightarrow r = 4 \\
 \]
Therefore the coefficient of \[{x^{32}}\]will be
\[{}^{15}{C_4} = \dfrac{{15!}}{{\left( {15 - 4} \right)!4!}} = \dfrac{{15!}}{{\left( {11} \right)!4!}} = \dfrac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2}} = 1365\]
Now for the term\[{x^{ - 17}}\], we can write
\[
\Rightarrow 60 - 7r = - 17 \\
\Rightarrow 60 + 17 = 7r \\
\Rightarrow r = 11 \\
 \]
Hence the coefficient of \[{x^{ - 17}}\] will be
\[{}^{15}{C_{11}} = - \dfrac{{15!}}{{\left( {15 - 11} \right)!11!}} = - \dfrac{{15!}}{{\left( 4 \right)!11!}} = - \dfrac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2}} = - 1365\]
Therefore the coefficient of \[{x^{32}} = 1365\] and \[{x^{ - 17}} = - 1365\].

Note: Students must note that \[{}^n{C_r}\] is the mathematical representation of the combination which is a method of selection of some items or all of the items from a set without taking the sequence of selection into consideration whereas in the case of permutation which is the method of arrangements of items of a set the sequence is considered represented as \[{}^n{P_r}\].