Question

find the coefficient of the term involving x¹⁰ in the expansion of (x²-2)¹¹




please help me
class 11th chepter no 9 binomial theorem​

Answer 1

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

Given expansion is

$\rm :\longmapsto\: {( {x}^{2} - 2) }^{11}$

We know that

$\red{ \sf{ \: The \: general \: term \: in \: the \: expansion \: of \: {(x + y)}^{n} \: is}}$

$\red{\rm :\longmapsto\:\boxed{\sf{ T_{r + 1} \: = \: ^{n} C_{r} \: {x}^{n - r} \: {y}^{r}}}}$

So, General term of the given expansion is

$\rm :\longmapsto\:\sf{ T_{r + 1} \: = \: ^{11} C_{r} \: { ({x}^{2} )}^{11 - r} \: {( - 2)}^{r}}$

$\rm :\longmapsto\:\sf{ T_{r + 1} \: = \: ^{11} C_{r} \: { ({x} )}^{22 - 2r} \: {( - 2)}^{r}}$

Now, we have to find the coefficient of x¹⁰,

So, Substitute

$\rm :\longmapsto\:22 - 2r = 10$

$\rm :\longmapsto\: - 2r = 10 - 22$

$\rm :\longmapsto\: - 2r = - 12$

$\bf\implies \:r = 6$

Thus,

$\rm :\longmapsto\:\sf{ T_{r + 1} \: = \: ^{11} C_{r} \: { ({x} )}^{22 - 2r} \: {( - 2)}^{r}}$

can be rewritten as on substituting r = 6

$\rm :\longmapsto\:\sf{ T_{6+ 1} \: = \: ^{11} C_{6} \: { ({x} )}^{10} \: {( - 2)}^{6}}$

$\rm :\longmapsto\:\sf{ T_{7} \: = \: ^{11} C_{6} \: { ({x} )}^{10} \: {2}^{6} }$

Hence,

$\rm :\longmapsto\:Coefficient \: of \: {x}^{10} = \: ^{11} C_{6} \times {2}^{6}$

$\rm :\longmapsto\:Coefficient \: of \: {x}^{10} = \: \dfrac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} \times 64$

$\rm :\longmapsto\:Coefficient \: of \: {x}^{10} = \: 462 \times 64$

$\rm :\longmapsto\:Coefficient \: of \: {x}^{10} = \: 29568$

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LEARN MORE

$\boxed{\tt{ ^{n} C_{r} = \: ^{n} C_{n - r}}}$

$\boxed{\tt{ ^{n} C_{0} = \: ^{n} C_{n} = 1}}$

$\boxed{\tt{ ^{n} C_{1} = \: ^{n} C_{n - 1} = n}}$

$\boxed{\tt{ ^{n} C_{x} = \: ^{n} C_{y}\rm\implies \:n = x + y \: \: or \: \: x = y}}$

$\boxed{\tt{ \: ^{n} C_{r} \: + \: ^{n} C_{r - 1} \: = \: ^{n + 1} C_{r}}}$

$\boxed{\tt{ \: \frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r} \: }}$

Answer 2

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

Given expansion is

$\rm :\longmapsto\: {( {x}^{2} - 2) }^{11}$

We know that

$\red{ \sf{ \: The \: general \: term \: in \: the \: expansion \: of \: {(x + y)}^{n} \: is}}$

$\red{\rm :\longmapsto\:\boxed{\sf{ T_{r + 1} \: = \: ^{n} C_{r} \: {x}^{n - r} \: {y}^{r}}}}$

So, General term of the given expansion is

$\rm :\longmapsto\:\sf{ T_{r + 1} \: = \: ^{11} C_{r} \: { ({x}^{2} )}^{11 - r} \: {( - 2)}^{r}}$

$\rm :\longmapsto\:\sf{ T_{r + 1} \: = \: ^{11} C_{r} \: { ({x} )}^{22 - 2r} \: {( - 2)}^{r}}$

Now, we have to find the coefficient of x¹⁰,

So, Substitute

$\rm :\longmapsto\:22 - 2r = 10$

$\rm :\longmapsto\: - 2r = 10 - 22$

$\rm :\longmapsto\: - 2r = - 12$

$\bf\implies \:r = 6$

Thus,

$\rm :\longmapsto\:\sf{ T_{r + 1} \: = \: ^{11} C_{r} \: { ({x} )}^{22 - 2r} \: {( - 2)}^{r}}$

can be rewritten as on substituting r = 6

$\rm :\longmapsto\:\sf{ T_{6+ 1} \: = \: ^{11} C_{6} \: { ({x} )}^{10} \: {( - 2)}^{6}}$

$\rm :\longmapsto\:\sf{ T_{7} \: = \: ^{11} C_{6} \: { ({x} )}^{10} \: {2}^{6} }$

Hence,

$\rm :\longmapsto\:Coefficient \: of \: {x}^{10} = \: ^{11} C_{6} \times {2}^{6}$

$\rm :\longmapsto\:Coefficient \: of \: {x}^{10} = \: \dfrac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} \times 64$

$\rm :\longmapsto\:Coefficient \: of \: {x}^{10} = \: 462 \times 64$

$\rm :\longmapsto\:Coefficient \: of \: {x}^{10} = \: 29568$

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LEARN MORE

$\boxed{\tt{ ^{n} C_{r} = \: ^{n} C_{n - r}}}$

$\boxed{\tt{ ^{n} C_{0} = \: ^{n} C_{n} = 1}}$

$\boxed{\tt{ ^{n} C_{1} = \: ^{n} C_{n - 1} = n}}$

$\boxed{\tt{ ^{n} C_{x} = \: ^{n} C_{y}\rm\implies \:n = x + y \: \: or \: \: x = y}}$

$\boxed{\tt{ \: ^{n} C_{r} \: + \: ^{n} C_{r - 1} \: = \: ^{n + 1} C_{r}}}$

$\boxed{\tt{ \: \frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r} \: }}$