Question

Find the middle term of (2x-3/x)^9

Answer 1

Step-by-step explanation:

We have,

$\bigg( 2x - \frac{3}{x} \bigg)^{9}$

Here, no. of terms is 10, so, there are two middle terms $T_{\frac{9+1}{2}}\:\:\&\:\:T_{\frac{9+3}{2}}$

So,

General term,

$T_{r+1} = \:^{9} C _{r}(2x)^{9 - r} . \bigg( \frac{3}{x} \bigg)^{r}$

$\implies \: T_{ \frac{9 + 1}{2} } = T_{ 5 } = T_{ 4 + 1} = \:^{9} C _{4}(2x)^{9 - 4} . \bigg( \frac{3}{x} \bigg)^{4}$

$\implies \: T_{ 5 } = \:^{9} C _{4}(2x)^{5} . \bigg( \frac{3}{x} \bigg)^{4}$

$\implies \: T_{ 5 } = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} .32(x)^{5} . \frac{81}{(x)^{4}}$

$\implies \: T_{ 5 } = \frac{9 \times 8 \times 7 \times 6 \times 32 \times 81}{4 \times 3 \times 2 \times 1} (x)$

$\implies \: T_{ 5 } = 326592x$

Also,

$\implies \: T_{ \frac{9 + 3}{2} } = T_{ 6 } = T_{ 5 + 1} = \:^{9} C _{5}(2x)^{9 - 5} . \bigg( \frac{3}{x} \bigg)^{5}$

$\implies \:T_{ 6 } = \:^{9} C _{5}(2x)^{4} . \bigg( \frac{3}{x} \bigg)^{5}$

$\implies \:T_{ 6 } = \:^{9} C _{5}. {2}^{4} (x)^{4} . \frac{3 ^{5} }{x^{5} }$

$\implies \:T_{ 6 } = \: \frac{^{9} C _{5}. {2}^{4}. {3}^{5} }{x }$