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

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Question

(c) The general term in the expansion of ${\left( {\frac{3}{2}{x^2} - \frac{1}{{3x}}} \right)^9}$ is ${T_{r + 1}} = {\,^9}{C_r}{\left( {\frac{3}{2}{x^

Vedclass · Accepted Answer

$\frac{{17}}{{54}}$

Vedclass · Answer

(c) The general term in the expansion of ${\left( {\frac{3}{2}{x^2} - \frac{1}{{3x}}} ight)^9}$ is ${T_{r + 1}} = {\,^9}{C_r}{\left( {\frac{3}{2}{x^2}} ight)^{9 - r}}{\left( { - \frac{1}{{3x}}} ight)^r}$$ = {\,^9}{C_r}{\left( {\frac{3}{2}} ight)^{9 - r}}\left( { - \frac{1}{3}} ight){x^{18 - 3r}}$
......$(i)$

Now, the coefficient of the term independent of $x$ in the expansion of $(1 + x + 2{x^3})\,\,{\left( {\frac{3}{2}{x^2} - \frac{1}{{3x}}} ight)^9}$ ......$(ii)$

= Sum of the coefficient of the terms ${x^0},{x^{ - 1}}$and ${x^{ - 3}}$ in ${\left( {\frac{3}{2}{x^2} - \frac{1}{{3x}}} ight)^9}$.

For ${x^0}$ in $(i)$ above, $18 - 3r = 0 \Rightarrow r = 6$. For ${x^{ - 1}}$ in

$(i)$ above, there exists no value of $r$ and hence no such term exists. For ${x^{ - 3}}$in 

$(i)$,$18 - 3r = - 3 \Rightarrow r = 7$

$	herefore $ For term independent of $x$, in $(ii)$ the coefficient

$ = 1 	imes {\,^9}{C_6}{( - 1)^6}{\left( {\frac{3}{2}} ight)^{9 - 6}}{\left( {\frac{1}{3}} ight)^6} + 2 	imes {\,^9}{C_7}{( - 1)^7}{\left( {\frac{3}{2}} ight)^{9 - 7}}{\left( {\frac{1}{3}} ight)^7}$

$ = \frac{{9.8.7}}{{1.2.3}}.\frac{{{3^3}}}{{{2^3}}}.\frac{1}{{{3^6}}} + 2\frac{{9.8}}{{1.2}}( - 1)\frac{{{3^2}}}{{{2^2}}}.\frac{1}{{{3^7}}}$$ = \frac{7}{{18}} - \frac{2}{{27}} = \frac{{17}}{{54}}$.

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

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