Question

please help me in this ​

Answer 1

To Find :-

$\sf{ \implies \: Term \: independent \: of \: x \: in \: ( {x}^{2} - \frac{1}{x} )^{9} }$

Solution :-

Here we have to use binomial theorem.

$\sf{ \implies \: Term \: independent \: of \: x \: is \: required }$

→ Let's suppose (r+1)th term contains x⁰ .

$\sf{ \implies \: T_{r + 1} = {}^{n}C _r .\: {x}^{n - r} .\: {a}^{r} }$

$\sf{ \implies \: T_{r + 1} = {}^{9}C _ r.\: {{x}^{2}}^{9 - r} .\: {\dfrac{1}{x}}^{r} }$

$\sf{ \implies \: T_{r + 1} = {}^{9}C _ r.\: {x}^{2 \times 9 - 2r }.\: {x}^{-r} }$

$\sf{ \implies \: T_{r + 1} = {}^{9}C _ r.\: {x}^{18 - 2r }.\: {x}^{-r} }$

$\sf{ \implies \: T_{r + 1} = {}^{9}C _ r.\: {x}^{18 -2r - r} }$

$\sf{ \implies \: T_{r + 1} = {}^{9}C _ r.\: {x}^{18 -3r} }$

$\sf{ \implies \: {x}^{0} = {x}^{18-3r} }$

$\sf{ \implies \: 18 - 3r = 0 }$

$\sf{ \implies \: 3r = 18 }$

$\sf{ \implies \: r = \dfrac{18}{3} = 6 }$

The term independent of x is 7th

So 6+1 = 7 th

Verification :-

$\sf{ \implies \: T_{6 + 1} = {}^{9}C _6 .\: {{x}^{2}}^{9 - 6} .\: {\dfrac{1}{x}}^{6} }$

$\sf{ \implies \: T_{6 + 1} = {}^{9}C _6 .\: {{x}^{2}}^{3} .\: {\dfrac{1}{x}}^{6} }$

$\sf{ \implies \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} \times {x}^{6} \times \dfrac{1}{ {x}^{6} } }$

$\sf{ \implies T_7 = \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} }$

$\sf{ \implies T_7 = 84 }$

Answer 2

Answer:

if I do this then my first account means this account will not get deleted na