Answer

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**Hint:**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.

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