Math, asked by ITZSnowyBoy, 13 days ago

 \huge \red{Question : - }

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

Answers

Answered by mathdude500
4

\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}  \: }}

Answered by OoAryanKingoO79
16

Answer:

your answer in the attachment

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