Question

The binomial coefficient of the third term of the end in the expansion of ${\left( {{y^{2/3}} + {x^{5/4}}} \right)^n}$ is 91. Find the 9th term of the expansion.

Answer 1

We will find the value of $n$ by equating 91 with the coefficient of the third last term of the expansion. We will find the ${9^{th}}$ term by substituting the value of $n$ that we have calculated in the formula for ${\left( {r + 1} \right)^{th}}$ term where $r$ is 8.

Formulas used:
We will use the following formulas:
$^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$, there are $n$ number of objects and $r$ selections.
In the expansion of ${\left( {a + b} \right)^n}$, the ${\left( {r + 1} \right)^{th}}$ term is given by ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$.
$n! = n\left( {n - 1} \right)\left( {n - 2} \right)...3 \cdot 2 \cdot 1$
Roots of the quadratic equation $a{x^2} + bx + c = 0$ are given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
${\left( {{p^a}} \right)^b} = {p^{ab}}$

Complete step by step answer:
We know that the expansion of ${\left( {a + b} \right)^n}$ has $n + 1$ terms. So, the 3rd term from the end will be the ${\left( {n - 1} \right)^{th}}$ term from the beginning.
To find the ${\left( {n - 1} \right)^{th}}$ term of the expansion, we will substitute $n - 2$ for $r$, ${y^{2/3}}$ for $a$ and ${x^{5/4}}$ for $b$ in the formula ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$. Therefore, we get
${T_{n - 2 + 1}} = {}^n{C_{n - 2}}{\left( {{y^{2/3}}} \right)^{n - \left( {n - 2} \right)}}{\left( {{x^{5/4}}} \right)^{n - 2}}$
The binomial coefficient of the 3rd term from the end is ${}^n{C_{n - 2}}$.
Now using the formula $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$, we get
$\begin{array}{l}{}^n{C_{n - 2}} = \dfrac{{n!}}{{\left( {n - \left( {n - 2} \right)} \right)!\left( {n - 2} \right)!}}\\\ \Rightarrow {}^n{C_{n - 2}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{2!\left( {n - 2} \right)!}}\\\ \Rightarrow {}^n{C_{n - 2}} = \dfrac{{n\left( {n - 1} \right)}}{2}\end{array}$
We also know that the binomial coefficient of the 3rd term from the end is 91. We will equate the 2 terms:
$\Rightarrow \dfrac{{n\left( {n - 1} \right)}}{2} = 91$
On cross multiplication, we get
$\begin{array}{l} \Rightarrow n\left( {n - 1} \right) = 91 \cdot 2\\\ \Rightarrow {n^2} - n - 182 = 0\end{array}$
The above equation is a quadratic equation.
Now using the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and simplifying, we get
$\begin{array}{l}n = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( { - 182} \right)} }}{2}\\\ \Rightarrow n = \dfrac{{1 \pm \sqrt {729} }}{2}\end{array}$
Simplifying further, we get
$\Rightarrow n = \dfrac{{1 \pm 27}}{2}$
$\Rightarrow n = 14$ or $n = - 13$
We will reject $- 13$ for $n$ as the number of terms cannot be negative. So, the value of $n$ is 14.
We will find the 9th term of the expansion. We will substitute 14 for $n$, 8 for $r$, ${y^{2/3}}$for $a$ and ${x^{5/4}}$ for $b$ in the formula ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$. Therefore, we get
${T_{8 + 1}} = {}^{14}{C_8}{\left( {{y^{2/3}}} \right)^6}{\left( {{x^{5/4}}} \right)^8}$
We will substitute 14 for $n$ and 8 for $r$ in the first formula and we will use the formula ${\left( {{p^a}} \right)^b} = {p^{ab}}$ to simplify the above equation:
$\begin{array}{l} \Rightarrow {T_9} = \dfrac{{14!}}{{\left( {14 - 8} \right)!8!}} \cdot {y^{\dfrac{2}{3} \times 6}} \cdot {x^{\dfrac{5}{4} \times 8}}\\\ \Rightarrow {T_9} = \dfrac{{14!}}{{6!8!}} \cdot {y^{2 \times 2}} \cdot {x^{5 \times 2}}\end{array}$
Applying the factorial, we get
$\Rightarrow {T_9} = 3003{y^4}{x^{10}}$

$\therefore$ The 9th term of the expansion is $3003{y^4}{x^{10}}$.

Note:

1. The coefficient of the ${r^{th}}$ term from the beginning and end of a binomial expansion is the same. For example,
   $\begin{array}{l}{}^n{C_0} = {}^n{C_n}\\\\{}^n{C_1} = {}^n{C_{n - 1}}\\\\{}^n{C_r} = {}^n{C_{n - r}}\end{array}$
2. We can use this property to find the coefficient of the 3rd term from the end.
   $\begin{array}{l}{}^n{C_2} = 91\\\ \Rightarrow \dfrac{{n!}}{{\left( {n - 2} \right)!2!}} = 91\end{array}$
3. It is the same as the coefficient of the 3rd term from the beginning.