(1 + x + 2x^3) {left( {frac{3}{2}{x^2} - frac | vedclass

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Question

The $(r+1)$ th term in the expansion of ${\left[ {(3/2){x^2} - (1/3x)} \right]^9}$ is given by

${T_{r + 1}} = {\,^9}{C_r}{\left( {\frac{3}{2}{x^2}}

Vedclass · Accepted Answer

$17/54$

Vedclass · Answer

The $(r+1)$ th term in the expansion of ${\left[ {(3/2){x^2} - (1/3x)} \right]^9}$ is given by

${T_{r + 1}} = {\,^9}{C_r}{\left( {\frac{3}{2}{x^2}} \right)^{9 - r}}{\left( { - \frac{1}{{3x}}} \right)^r} = {\,^9}{C_r}{( - 1)^r}\frac{{{3^{9 - 2r}}}}{{{2^{9 - r}}}}{x^{18 - 3r}}$      .....$(1)$

Since we are looking for the coefficient of the term independent of $x$ in the expansion of

$\left(1+x+2 x^{3}\right)\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$         .....$(2)$

we must get the coefficient of $x^{0}, x^{-1}$ and $x^{-3}$ in the expansion of ${\left[ {(3/2){x^2} - (1/3x)} \right]^9}.$

For $x^{0}, r$ must be $6$ in $( 1);$ for $x^{-1},$ there is no value of $r ;$ and for $x^{-3}, r$ must be $7$

in $(1) .$ Therefore, the coefficient of the term independent of $x$ in $( 2)$ is

$1.{\,^9}{{\rm{C}}_6}{( - 1)^6} \cdot \frac{{{3^{9 - 12}}}}{{{2^{9 - 6}}}} + 2 \cdot {\,^9}{{\rm{C}}_7}{( - 1)^7} \cdot \frac{{{3^{9 - 14}}}}{{{2^{9 - 7}}}}$

${=\frac{9.8 .7}{1.2 .3} \cdot \frac{3^{-3}}{2^{3}}+2 \frac{9.8}{1.2}(-1) \cdot \frac{3^{-5}}{2^{2}}=\frac{7}{18}-\frac{2}{27}=\frac{17}{54}}$

$(1 + x + 2x^3)$ ${\left( {\frac{3}{2}{x^2} - \frac{1}{{3x}}} \right)^9}$ ના વિસ્તરણમાં અચળ પદ મેળવો

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