Question

༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅᴇʀᴀᴛᴇʀs

༒Other best user Only

Find the term independent of x in the expansion of
$( \sqrt{ \frac{x}{3} } - \frac{ \sqrt{3} }{2x} ) {}^{12}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expansion is

$\rm :\longmapsto\:\bigg[ \sqrt{ \dfrac{x}{3} } - \dfrac{ \sqrt{3} }{2x}\bigg]^{12}$

can be rewritten as

$\rm \: = \: {\bigg[\dfrac{ \sqrt{x} }{ \sqrt{3} } + \dfrac{ \sqrt{3} }{2x} \bigg]}^{12}$

We know,

In Binomial expansion,

$\rm :\longmapsto\: {(x + y)}^{n}$

The general term is given by

$\boxed{ \bf \: T_{r+1} \: = \: ^nC_r \: {x}^{n - r} \: {y}^{r}}$

So,

General Term of given expansion is

$\rm :\longmapsto\:T_{r+1} = \: ^{12}C_r \: {\bigg[\dfrac{ \sqrt{x} }{ \sqrt{3} } \bigg]}^{12 - r} \: {\bigg[\dfrac{ \sqrt{3} }{2x} \bigg]}^{r}$

$\rm \: = \: ^{12}C_r \: \dfrac{ {\bigg[x\bigg]}^{\dfrac{12 - r}{2} } }{{\bigg[3\bigg]}^{\dfrac{12 - r}{2} }} \: \dfrac{ {( \sqrt{3} )}^{r} }{ {2}^{r} {x}^{r} }$

$\rm \: = \: ^{12}C_r \: \dfrac{ {\bigg[x\bigg]}^{\dfrac{12 - r}{2} - r} }{{\bigg[3\bigg]}^{\dfrac{12 - r}{2} }} \: \dfrac{ {( \sqrt{3} )}^{r} }{ {2}^{r} }$

$\rm \: = \: ^{12}C_r \: \dfrac{ {\bigg[x\bigg]}^{\dfrac{12 - r - 2r}{2}} }{{\bigg[3\bigg]}^{\dfrac{12 - r}{2} }} \: \dfrac{ {( \sqrt{3} )}^{r} }{ {2}^{r} }$

$\rm \: = \: ^{12}C_r \: \dfrac{ {\bigg[x\bigg]}^{\dfrac{12-3 r }{2}} }{{\bigg[3\bigg]}^{\dfrac{12 - r}{2} }} \: \dfrac{ {( \sqrt{3} )}^{r} }{ {2}^{r} }$

So,

$\rm :\longmapsto\:T_{r+1} \rm \: = \: ^{12}C_r \: \dfrac{ {\bigg[x\bigg]}^{\dfrac{12- 3r }{2}} }{{\bigg[3\bigg]}^{\dfrac{12 - r}{2} }} \: \dfrac{ {( \sqrt{3} )}^{r} }{ {2}^{r} }$

For the term to be independent of x, we have to substitute

$\rm :\longmapsto\:\dfrac{12-3 r}{2} = 0$

$\rm :\longmapsto\:12 - 3r = 0$

$\bf\implies \:r = 4$

Hence,

$\bf\implies \:T_{5} \: term \: is \: independent \: of \: x$

So, Required term is

$\rm :\longmapsto\:T_{5} \rm \: = \: ^{12}C_{4}\: \dfrac{ 1 }{{\bigg[3\bigg]}^{\dfrac{12 - 4}{2} }} \: \dfrac{ {( \sqrt{3} )}^{4} }{ {2}^{4} }$

$\rm :\longmapsto\:T_{5} \rm \: = \: ^{12}C_{4}\: \dfrac{ 1 }{{\bigg[3\bigg]}^{\dfrac{8}{2} }} \: \dfrac{ {3}^{2} }{ {2}^{4} }$

$\rm :\longmapsto\:T_{5} \rm \: = \: \dfrac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \: \times \dfrac{ 1 }{{ {3}^{4} }} \times \: \dfrac{ {3}^{2} }{ {2}^{4} }$

$\rm :\longmapsto\:T_{5} \rm \: = 495 \times \dfrac{1}{9} \times \dfrac{1}{16}$

$\rm :\longmapsto\:T_{5} \rm \: = 55 \times \dfrac{1}{16}$

$\rm :\longmapsto\:T_{5} \rm \: = \dfrac{55}{16}$

Additional Information :-

$\boxed{ \sf \: ^nC_1 = ^nC_{n - 1} = n}$

$\boxed{ \sf \: ^nC_0 = ^nC_{n} = 1}$

$\boxed{ \sf \: \frac{^nC_r}{^nC_{r - 1}} = \frac{n - r + 1}{r}}$

$\boxed{ \sf \: ^nC_r \: + \: ^nC_{r - 1} \: = \: ^{n + 1}C_r}$