Question

Find the term independent of x in the expansion of
$( \sqrt{ \frac{x}{3} } - \frac{ \sqrt{3} }{2x} ) {}^{12}$
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Answer 1

Step-by-step explanation:

We have,

$\bigg(\sqrt{ \frac{x}{3} } - \frac{ \sqrt{3} }{2x} \bigg)^{12}$

The general term on expanding this is,

$T_{r + 1} = \sum_{r = 0}^{12} ( - 1)^{r}. ^{12}C_{r} \bigg( \sqrt{ \frac{x}{3} } \bigg) ^{12 - r}. \bigg( \frac{ \sqrt{3} }{2x} \bigg) ^{r}$

$\implies \: T_{r + 1} = ( - 1)^{r}. ^{12}C_{r} \bigg( \frac{x}{3} \bigg) ^{ \frac{12 - r}{2}}. \bigg( \frac{ \sqrt{3} }{2x} \bigg) ^{r}$

$\implies \: T_{r + 1} = ( - 1)^{r}. ^{12}C_{r} \bigg( \frac{x ^{6 - \frac{ r}{2}} }{3^{6 - \frac{ r}{2}} } \bigg) . \bigg( \frac{ 3^{ \frac{r}{2} } }{2^{r}. x^{r} } \bigg)$

$\implies \: T_{r + 1} = ( - 1)^{r}. ^{12}C_{r} ( x ^{6 - \frac{ r}{2} - r} ) . \bigg( \frac{ 3^{ \frac{r}{2} - 6 + \frac{r}{2} } }{2^{r} } \bigg)$

$\implies \: T_{r + 1} = ( - 1)^{r}. ^{12}C_{r} ( x )^{6 - \frac{ 3r}{2} } . \bigg( \frac{ 3^{ r- 6 } }{2^{r} } \bigg)$

Now, for term independent of x, $6 - \frac{ 3r}{2}=0$

$\implies \: 6 = \frac{3 r}{2}$

$\implies \: r = 4$

So,

$\implies \: T_{4 + 1} = ( - 1)^{4}. ^{12}C_{4} ( x )^{6 - \frac{ 3 \times 4}{2} } . \bigg( \frac{ 3^{ 4- 6 } }{2^{4} } \bigg)$

$\implies \: T_{5} = ( 1) . \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} .( x )^{0} . \bigg( \frac{ 3^{ - 2 } }{2^{4} } \bigg)$

$\implies \: T_{5} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} . \bigg( \frac{1 }{2 \times 2 \times 2 \times 2 \times 3 \times 3 } \bigg)$

$\implies \: T_{5} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} . \bigg( \frac{1 }{16 \times 9 } \bigg)$

$\implies \: T_{5} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1 \times 16 \times 9}$

$\implies \: T_{5} = \frac{12 \times 11 \times 10 }{4 \times 3 \times 2 \times 1 \times 16 }$

$\implies \: T_{5} = \frac{11 \times 10 }{ 2 \times 1 \times 16 }$

$\implies \: T_{5} = \frac{11 \times 5 }{ 1 \times 16 }$

$\implies \: T_{5} = \frac{55 }{ 16 }$