Question

12
go) Find the middle terms in the expanision (x/y+y/x) 12​

Answer 1

Basic Concept :-

$\sf\: Let \: us \: consider \: the \: expansion \: of \: {(x + a)}^{n} \: then$

$\sf\: middle \: term \: depends \: on \: value \: of \: n.$

Two cases arises :-

Case 1 :- If n is even, then there is only middle term, i.e.

$\rm :\longmapsto\:T_{\frac{n + 2}{2}} \: term$

Case 2 :- If n is odd, then there are 2 middle terms.

$\rm :\longmapsto\:T_{\frac{n + 1}{2}} \:and \: T_{\frac{n + 3}{2}} \: term$

and

General term of Binomial expansion is given by

$\rm :\longmapsto\:T_{r + 1} \: = \: \:^n C_r \: {x}^{n - r} \: {a}^{r}$

and

$\boxed{\sf \:^{n}C_r=\dfrac{n!}{(n-r)!\times r!}}$

Let's solve the problem now!!!

Given expansion is

$\green{\rm :\longmapsto\: \bf{ \: {\bigg(\dfrac{x}{y} + \dfrac{y}{x}\bigg) }^{12}}}$

Here, n = 12

which is even,

This implies, there is only 1 middle term, i.e

$\blue{\rm :\longmapsto\: \bf \:T_{\dfrac{12 + 2}{2}} = T_7}$

Now,

$\rm :\longmapsto\:T_7$

$\rm \: = \: \: \:T_{6 + 1}$

$\rm \: = \: \: \:^{12} C_6 \: {\bigg(\dfrac{x}{y}\bigg) }^{12 - 6}{\bigg(\dfrac{y}{x}\bigg) }^{6}$

$\rm \: = \: \: \:^{12} C_6 \: {\bigg(\dfrac{x}{y}\bigg) }^{6}{\bigg(\dfrac{y}{x}\bigg) }^{6}$

$\rm \: = \: \: \:^{12} C_6 \: {\bigg(\dfrac{x}{y}\bigg) }^{6}{\bigg(\dfrac{x}{y}\bigg) }^{ - 6}$

$\rm \: = \: \: \:^{12} C_6 \:$

$\rm \: = \: \: \dfrac{12!}{(12-6)!\times 6!}$

$\rm \: = \: \: \dfrac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times \cancel {6!}}{6!\times \cancel{6!}}$

$\rm \: = \: \: \dfrac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$\rm \: = \: \: 924$

Hence,

$\green{\rm :\longmapsto\: \bf{ \:Middle \: term \: in \: {\bigg(\dfrac{x}{y} + \dfrac{y}{x}\bigg) }^{12} \: = \: 924}}$

Additional Information :-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \: \:^n C_r + \:^n C_{r - 1} \: = \:^{n + 1} C_r}}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\blue{{\tt \: \dfrac{\:^n C_r}{\:^n C_{r - 1}} = \dfrac{n - r + 1}{r} }}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\blue{{\tt \: \:^n C_x = \:^n C_y \implies \: x = y \: or \: n = x + y}}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\blue{{\tt \: \:^n C_0 = \:^n C_n \: = \: 1 }}}}}} \\ \end{gathered}$