The term independent of x in the expansion of | vedclass

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(A) The general term in the expansion of ${\left( {{x^2} - \frac{{3\sqrt{3}}}{{{x^3}}}} \right)^{10}}$ is given by: ${T_{r + 1}} = {}^{10}{C_r} {({x^2

Vedclass · Accepted Answer

$153090$

Vedclass · Answer

(A) The general term in the expansion of ${\left( {{x^2} - \frac{{3\sqrt{3}}}{{{x^3}}}} ight)^{10}}$ is given by: ${T_{r + 1}} = {}^{10}{C_r} {({x^2})^{10 - r}} {\left( { - 3\sqrt{3} \cdot {x^{ - 3}}} ight)^r}$ ${T_{r + 1}} = {}^{10}{C_r} {x^{20 - 2r}} {(-3\sqrt{3})^r} {x^{ - 3r}}$ ${T_{r + 1}} = {}^{10}{C_r} {(-3\sqrt{3})^r} {x^{20 - 5r}}$ For the term to be independent of $x$,the exponent of $x$ must be $0$: $20 - 5r = 0 \Rightarrow r = 4$ Substituting $r = 4$ into the expression: ${T_5} = {}^{10}{C_4} {(-3\sqrt{3})^4}$ ${T_5} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} 	imes {(-3)^4} 	imes {(\sqrt{3})^4}$ ${T_5} = 210 	imes 81 	imes 9$ ${T_5} = 210 	imes 729 = 153090$

The term independent of $x$ in the expansion of ${\left( {{x^2} - \frac{{3\sqrt{3}}}{{{x^3}}}} \right)^{10}}$ is

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