Question

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Answer 1

${\huge{\boxed{\boxed{\underline{Answer}}}}}$

Option C).

Answer 2

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

Given binomial expression is

$\rm :\longmapsto\: {\bigg( {x}^{2} - \dfrac{1}{2x} \bigg) }^{20}$

It is given that rth term is the middle term of the given expansion.

Here,

• n = 20

It means, there is only 1 middle term, i.e. 11 th term

So, it means rth term is 11th term.

As, we have to find (r + 3)th term.

It means, we have to find 14th term.

We know,

$\rm :\longmapsto\:In \: binomial \:expansion \: of\: {(x + y)}^{n}, \:$

$\boxed{\tt{ \: The \: general \: term \: is \: T_{r + 1} \: = \: ^nC_{r} \: {x}^{n - r} \: {y}^{r} \: }}$

So,

$\rm :\longmapsto\:T_{14}$

$\rm \: = \: T_{13 + 1}$

$\rm \: = \: ^{20}C_{13} {( {x}^{2} )}^{20 - 13} {\bigg( - \dfrac{1}{2x} \bigg) }^{13}$

$\rm \: = \: - \: ^{20}C_{13} {( {x}^{2} )}^{7} {\bigg( \dfrac{1}{2x} \bigg) }^{13}$

$\rm \: = \: - \: ^{20}C_{13} \times {x}^{14} \times \dfrac{1}{ {2}^{13} \: {x}^{13} }$

$\rm \: = \: - \: ^{20}C_{13} \times {x}^{14 - 13} \times \dfrac{1}{ {2}^{13} }$

$\rm \: = \: - \: ^{20}C_{13} \times {x}^{1} \times \dfrac{1}{ {2}^{13} }$

We know,

$\boxed{\tt{ \: ^nC_{r} \: = \: ^nC_{n - r} \: }}$

So, using this, we get

$\rm \: = \: - \: ^{20}C_{20 - 13} \times {x}^{1} \times \dfrac{1}{ {2}^{13} }$

$\rm \: = \: - \: ^{20}C_{7} \times x \times \dfrac{1}{ {2}^{13} }$

$\rm \: = \: - \: ^{20}C_{7} \: {2}^{ - 13} \: x$

So,

$\sf\implies \: \boxed{\tt{ \: T_{14} = \: - \: ^{20}C_{7} \: x \times \: {2}^{ - 13} \: }}$

Hence,

• Option (c) is correct.

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EXPLORE MORE

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

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

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

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

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