Question

which term in the expansion of
$\bigg( {x}^{2} - \dfrac{1}{x} \bigg)^{9}$
will be independent of heads also find that term ​

Answer 1

⇝ Given :-

A Binomial Expression of Index 9 :

${\bigg( {\text x}^{2} - \dfrac{1}{\text x} \bigg)}^{9}$

⇝ To Find :-

• The term independent of x in its Binomial expansion.

⇝ Main Concept :-

If n is any natural number and a , b are any numbers , then in the expansion of $\sf(a + b)^n$ (p + 1)th term is given by : .

$\large\pink{ \bf T_{p+1}={}^{p} C_{r}.{( a)}^{n - p}.( {b)}^{p}}$

⇝ Solution :-

Let (r + 1)th Term be the Term independent of x.

As We Know,

$\large\bf T_{r+1}={}^{n} C_{r}. {( a)}^{n- r}.( {b)}^{r}$

$\large = \sf{}^{9} C_{r}. ( {{x}^{2})}^{9 - r} . \bigg(- \frac{ 1}{x} \bigg)^{r}$

$= \sf{}^{9} C_{r}.( - 1)^{r}.({{x)}^{2(9 - r)}} . \bigg(\frac{ 1}{x} \bigg)^{r}$

$= \sf{}^{9} C_{r}.( - 1)^{r}.({{x)}^{(18 - 2r)}} . \bigg(\frac{1}{x ^{r} } \bigg)$

$= \sf{}^{9} C_{r}.( - 1)^{r}.({{x)}^{(18 - 2r - r)}}$

$\red{:\longmapsto\bf T_{r+1}={}^{9} C_{r}.( - 1)^{r}.({{x)}^{(18 - 3r)}}}$

★ As it is independent of x so power of x should be zero.

$:\longmapsto18 - 3\text r = 0$

$:\longmapsto3\text r = 18$

$:\longmapsto\text r = \cancel\dfrac{18}{3}$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\bf r = 6} }}}$

★ According to our Assumption

(r + 1)th term is independent of x,

Hence,

$\bold{{Term \: independent \: of \: x =(6 + 1)th }}$

So,

$\large\underline{\pink{\underline{\frak{\pmb{\text Term \: independent \: of \: x =7th }}}}}$

❒ Finding 7th Term :-

$\sf T_{6+1}={}^{9} C_{6}.( - 1)^{6}.({{x)}^{(18 - 3 \times 6)}}$

$= \sf {}^{9} C_{6} \times 1$

$= \sf{}^{9} C_{6}$

$\sf = \dfrac{9!}{(9 - 6)! \times 6!}$

$= \frac{9!}{3! \times 6!}$

$= \frac{9 \times 8 \times 7 \times \cancel{6!}}{3 \times 2 \times \cancel{6 !}}$

$= 84$

Hence,

$\Large\underline{\pink{\underline{\frak{\pmb{\text 7th \: term=84 }}}}}$

Answer 2

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