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Question

The constant term in the expansion of $$ \left( x^2 + \dfrac {1}{x^2} + y + \dfrac {1}{y} \right)^8 $$ is :


A
4900
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B
4950
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C
5050
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D
5151
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Solution

The correct option is C $$4900$$
We have,
$$\dfrac { { 8! } }{ { \alpha !\beta !\gamma !\delta ! } } { \left( { { x^{ 2 } } } \right) ^{ \alpha  } }{ \left( { \dfrac { 1 }{ { { x^{ 2 } } } }  } \right) ^{ \beta  } }{ \left( y \right) ^{ \gamma  } }{ \left( { \dfrac { 1 }{ y }  } \right) ^{ \delta  } } \\ \dfrac { { 8! } }{ { \alpha !\beta !\gamma !\delta ! } } { \left( x \right) ^{ 2\alpha -2\beta  } }\times { \left( y \right) ^{ \gamma -\delta  } }$$

$$ \\ For\, constant\, term\,  \\ \alpha =\beta \, \, ;\, \, \, \gamma =\delta  \\ \alpha +\beta \, +\, \gamma +\delta =8 \\ 2\alpha +2\, \gamma =8 \\ \alpha +\gamma =4 \\ \left( { \beta =1 } \right) \, \alpha =1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \gamma =3\, \left( { \delta =3 } \right) \, \,  \\ \left( { \beta =2 } \right) \, \, \alpha =2\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \gamma =2\, \left( { \delta =2 } \right)  \\ \left( { \beta =3 } \right) \, \alpha =3\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \gamma =1\, \left( { \delta =1 } \right)  \\ \left( { \beta =4 } \right) \, \alpha =4\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \gamma =0\, \left( { \delta =0 } \right)  \\ \left( { \beta =0 } \right) \, \alpha =0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \gamma =4\, \left( { \delta =4 } \right) \, $$

$$ \\ Cons\tan  t\, term\, =\dfrac { { 8! } }{ { 1!1!3!3! } } +\dfrac { { 8! } }{ { 2!2!2!2! } } +\dfrac { { 8! } }{ { 3!3!1!1! } } +\dfrac { { 8! } }{ { 4!4! } }  \\ =2\times \dfrac { { 8\times 7\times 6\times 5\times 4 } }{ 6 } +\dfrac { { 8\times 7\times 6\times 120 } }{ { 16 } } +\dfrac { { 8\times 7\times 6\times 5\times 4! } }{ { 24 } } \times 2 \\ =2240+2520+4\times 7\times 6 \\ =2240+2520+140 \\ =4900$$

$$ \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer.$$

Mathematics

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