Question

Find the middle term in the expansion of (x −3/x)^10

Answer 1

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

Formula Used :-

$\rm :\longmapsto\:In \: the \: expansion \: of \: {(x + a)}^{n}, \: general \: term \: is$

$\red{ \boxed{ \sf{T_{r + 1} \: = \: ^nC_r \: {(x)}^{n - r} \: {a}^{r} }}}$

Middle Term :-

Middle term depends on value of n.

If n is even, there is only one middle term, i.e.

$\rm :\longmapsto\:{\bigg(\dfrac{n}{2} + 1 \bigg) }^{th} \: term$

And

If n is odd, there are two middle terms, i.e.

$\rm :\longmapsto\:{\bigg(\dfrac{n + 1}{2} \bigg) }^{th} \: term \: and \: {\bigg(\dfrac{n + 3}{2} \bigg) }^{th} \: term$

Let's solve the problem now!!!

Given expansion is

$\rm :\longmapsto\: {\bigg(x - \dfrac{3}{x} \bigg) }^{10}$

Here,

n = 10 which is even.

Hence, there is only one middle term, i. e. 6th term.

So, middle term is evaluated as

$\rm :\longmapsto\:T_{5 + 1} = \: ^{10}C_5 \: {(x)}^{10 - 5}{\bigg(\dfrac{ - 3}{x} \bigg) }^{5}$

$\rm :\longmapsto\:T_{6} = \: \dfrac{10! }{5! \: 5! } \: {(x)}^{5}{\bigg(\dfrac{ - 3}{x} \bigg) }^{5}$

$\rm :\longmapsto\:T_{6} = \: \dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5! }{5 \times 4 \times 3 \times 2 \times 1 \times \: 5! } \: {( - 3)}^{5}$

$\rm :\longmapsto\:T_{6} = - 61236$

Additional Information :-

$\red{ \boxed{ \sf{ ^nC_0 = ^nC_n = 1\: }}}$

$\red{ \boxed{ \sf{ ^nC_1 = ^nC_{n - 1} = n\: }}}$

$\red{ \boxed{ \sf{^nC_r \: = \: \frac{n! }{r! \: (n - r)! } }}}$

$\red{ \boxed{ \sf{ ^nC_r\: = \: \frac{n}{r} \: ^{n - 1}C_{r - 1}}}}$

$\red{ \boxed{ \sf{^nC_r \: + \: ^{n}C_{r - 1} = \: ^{n + 1}C_{r}}}}$

$\red{ \boxed{ \sf{ \frac{^nC_r}{^{n }C_{r - 1}} \: } = \frac{n - r + 1}{r} }}$

$\red{ \boxed{ \sf{ The \: coefficient \: of \: {x}^{r} \: in \: {(1 + x)}^{n} \: is \: \: ^nC_r}}}$

$\red{ \boxed{ \sf{ The \: coefficient \: of \: T_{r + 1} \: in \: {(1 + x)}^{n} \: is \: \: ^nC_r}}}$