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 ${}^n{P_r}$.