Answer
Verified
387.9k+ views
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.
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.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE