Question

find the independent of x in expansion of (
$(x {}^{2} + \frac{1}{x} ) {}^{9}$
please answer for 5marks..​

Answer 1

Step-by-step explanation:

We have,

$( {x}^{2} + \frac{1}{x} )^{9}$

$let \: \: t _{r + 1} \: \: be \: \: the \: \: term \: \: independent \: \: of \: \: x$

$t_{r + 1} = \:^{9} c_{r}.( {x}^{2} )^{9 - r} .( \frac{1}{x} )^{r}$

$= \:^{9} c_{r}. {x}^{18 - 2r} . {x}^{ - r}$

$= \:^{9} c_{r}. {x}^{18 - 3r}$

Now, coefficient of x will be 0

so,

$18 - 3r = 0$

$\implies \: r = 6$

So, 7th term will the required term independent of x

$t _{7} = \:^{9} c_{6}. {x}^{0}$

$\implies \: t_{7} = \:^{9} c_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$