Question

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

Answer 1

Hint: First we’ll find the value of n using the binomial coefficient of the third term of the end which is given 91. After finding the value of n we’ll easily get the value of the ninth term using the formula of ${(r + 1)^{th}}$ term of the expansion of ${\left( {a + b} \right)^n}$ i.e. given by ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$

Complete step by step answer:

Given data: In the expansion of ${\left( {{y^{\dfrac{2}{3}}} + {x^{\dfrac{5}{4}}}} \right)^n}$, the binomial coefficient of the third term from last is 91.
We know that in the expansion if ${\left( {a + b} \right)^n}$,
Number of terms=n+1
The formula for ${(r + 1)^{th}}$ term i.e.(${T_{r + 1}}$) is given by
${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$
Since total terms in the expansion of${\left( {{y^{\dfrac{2}{3}}} + {x^{\dfrac{5}{4}}}} \right)^n}$ will be (n+1)
Therefore, the third term from the last will be ${\left( {n + 1 - 3 + 1} \right)^{th}}$ i.e. ${\left( {n - 1} \right)^{th}}$
It is given that the binomial coefficient ${\left( {n - 1} \right)^{th}}$ is equal to 91,
Since ${T_{n - 1}} = {}^n{C_{n - 2}}{\left( {{y^{\dfrac{2}{3}}}} \right)^{n - n + 2}}{\left( {{x^{\dfrac{5}{4}}}} \right)^{n - 2}}$
Therefore, the binomial coefficient of ${T_{n - 1}} = {}^n{C_{n - 2}}$
$\therefore {}^n{C_{n - 2}} = 91$
Using ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ ,
 $ \Rightarrow \dfrac{{n!}}{{\left( {n - n + 2} \right)!\left( {n - 2} \right)!}} = 91$
On simplifying we get,
$ \Rightarrow \dfrac{{n!}}{{\left( 2 \right)!\left( {n - 2} \right)!}} = 91$
As, \[2! = 2\], we get,
$ \Rightarrow \dfrac{{n!}}{{2\left( {n - 2} \right)!}} = 91$
On multiplying the entire equation with 2 we get,
$ \Rightarrow \dfrac{{n!}}{{\left( {n - 2} \right)!}} = 91(2)$
Using $n! = n(n - 1)(n - 2)!$, and factorizing the right side we get,
$ \Rightarrow \dfrac{{n(n - 1)(n - 2)!}}{{\left( {n - 2} \right)!}} = 13(7)(2)$
On cancelling common terms we get,
$ \Rightarrow n(n - 1) = (14)13$
On comparing the left-hand side and right-hand side of the above equation, we can say that,
$n = 14$
Now, using ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$
Substituting r=8, we’ll get the required term i.e. ${T_9}$
\[\therefore {T_9} = {}^{14}{C_8}{\left( {{y^{\dfrac{2}{3}}}} \right)^{14 - 8}}{\left( {{x^{\dfrac{5}{4}}}} \right)^8}\]
On simplifying we get,
\[ = {}^{14}{C_8}{\left( y \right)^{6\left( {\dfrac{2}{3}} \right)}}{\left( x \right)^{8\left( {\dfrac{5}{4}} \right)}}\]
\[ = {}^{14}{C_8}{y^4}{x^{10}}\]
Using ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ , we get
\[\therefore {T_9} = \dfrac{{14!}}{{8!\left( {14 - 8} \right)!}}{y^4}{x^{10}}\]
\[ = \dfrac{{14!}}{{8!6!}}{y^4}{x^{10}}\]
Now, expanding the value of 14! ,
\[ = \dfrac{{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8!}}{{8!6!}}{y^4}{x^{10}}\]
On cancelling common terms we get,
\[ = \dfrac{{14 \times 13 \times 12 \times 11 \times 10 \times 9}}{{6!}}{y^4}{x^{10}}\]
Expanding 6!, we get,
\[ = \dfrac{{14 \times 13 \times 12 \times 11 \times 10 \times 9}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{y^4}{x^{10}}\]
On simplifying we get,
\[ = 3003{y^4}{x^{10}}\]
Therefore the ninth term is \[3003{y^4}{x^{10}}\]

Note: In the expansion of a binomial function let say ${\left( {a + b} \right)^n}$, we can say that the binomial coefficient or ${r^{th}}$ is given by ${}^n{C_{r - 1}}$ . Therefore, according to the given data, we can say that the binomial coefficient of the third term from last is 91 i.e. ${}^n{C_{n - 2}} = 91$that is equivalent to the above solution.