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Mathematics Question on Binomial theorem

The term independent of xx in the expansion of (x3+32x2)10{{\left( \sqrt{\frac{x}{3}}+\frac{3}{2{{x}^{2}}} \right)}^{10}} is

A

54\frac{5}{4}

B

74\frac{7}{4}

C

94\frac{9}{4}

D

4545

Answer

54\frac{5}{4}

Explanation

Solution

General term, Tr+1=10Cr(x3)10−r.(32x2)r{{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{\left( \sqrt{\frac{x}{3}} \right)}^{10-r}}.{{\left( \frac{3}{2{{x}^{2}}} \right)}^{r}}
=10Cr3r−10−r2(12)r.x10−r2−2r{{=}^{10}}{{C}_{r}}{{3}^{r-\frac{10-r}{2}}}{{\left( \frac{1}{2} \right)}^{r}}.{{x}^{\frac{10-r}{2}-2r}}
=10Cr3r−3r−102(12)r.x10−5r2{{=}^{10}}{{C}_{r}}{{3}^{r-\frac{3r-10}{2}}}{{\left( \frac{1}{2} \right)}^{r}}.{{x}^{\frac{10-5r}{2}}}
For the term independent of x, put
10−5r2=0\frac{10-5r}{2}=0
⇒\Rightarrow 5r=105r=10
⇒\Rightarrow r=2r=2
∴\therefore The term independent of x,
T3=10C2 36−102(12)2{{T}_{3}}{{=}^{10}}{{C}_{2}}\,{{3}^{\frac{6-10}{2}}}{{\left( \frac{1}{2} \right)}^{2}} =10×92×32×4=54=\frac{10\times 9}{2\times {{3}^{2}}\times 4}=\frac{5}{4}