Question

$\huge \red{Question : - }$

Find the 4th term from the end in the expansion of $( \frac{x {}^{3} }{2} - \frac{2}{x {}^{2} } ) {}^{9}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given binomial expansion is

$\rm :\longmapsto\: {\bigg[\dfrac{ {x}^{3} }{2} - \dfrac{2}{ {x}^{2} } \bigg]}^{9}$

We know, in binomial expansion, nth term from the end is given by

$\boxed{ \sf{T_{r + 1} \: from \: end \: in \: {(x + y)}^{n} \: = \: ^{n}C_{r} \: {x}^{r} \: {y}^{n - r}}}$

So, using this result, we get

$\red{\rm :\longmapsto\:T_{4} \: from \: end \: }$

$\rm \:  =  \: T_{3 + 1} \: from \: end$

$\rm \:  =  \: ^{9}C_{3} \: {\bigg[\dfrac{ {x}^{3} }{2} \bigg]}^{3} {\bigg[-\dfrac{2}{ {x}^{2} } \bigg]}^{9 - 3}$

$\rm \:  =  \: \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} \times \: {\bigg[\dfrac{ {x}^{3} }{2} \bigg]}^{3} {\bigg[-\dfrac{2}{ {x}^{2} } \bigg]}^{6}$

$\rm \:  =  \: 84 \times \dfrac{ {x}^{9} }{8} \times \dfrac{64}{ {x}^{12} }$

$\rm \:  =  \: \dfrac{692}{ {x}^{3} }$

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More to Know :-

$\boxed{ \tt{ \: ^{n}C_{r} \: + \: ^{n}C_{r - 1} \: = \: ^{n + 1}C_{r} \: }}$

$\boxed{ \tt{ \: \frac{^{n}C_{r}}{^{n}C_{r - 1}} = \frac{n - r + 1}{r} \: }}$

$\boxed{ \tt{ \: ^{n}C_{x} = ^{n}C_{y} \: \rm \implies\:x = y \: \: or \: \: n = x + y \: }}$

$\boxed{ \tt{ \: ^{n}C_{0} \: = \: ^{n}C_{n} \: = \: 1 \: }}$

$\boxed{ \tt{ \: ^{n}C_{1} \: = \: ^{n}C_{n - 1} \: = \: n \: }}$

$\boxed{ \tt{ \: ^{n}C_{2} \: = \: ^{n}C_{n - 2} \: = \: \frac{n(n - 1)}{2} \: }}$

$\boxed{ \tt{ \: ^{n}C_{3} \: = \: ^{n}C_{n - 3} \: = \: \frac{n(n - 1)(n - 2)}{3.2.1} \: }}$

Answer 2

Answer:

your answer in the attachment