If $(2 x^{2} - x - 1)^{5} = a_{0} + a_{1} x + a_{2} x^{2} . . . . . . . . . . a_{10} x^{10}$ then $a_{2} + a_{4} + a_{6} + a_{8} + a_{10}$ =

__

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Solution

Put x = 1,

⇒ $(2 - 1 - 1)^{5} = a_{0} + a_{1} + a_{2} . . . . . . . . . . a_{10}$

⇒ $a_{0} + a_{1} + a_{2} + a_{3} + a_{4} . . . . . . . . . . a_{9} + a_{10} = 0$ ...........(1)

Put x = -1

⇒ $(2 + 1 - 1)^{5} = a_{0} - a_{1} + a_{2} - a_{3} . . . . . . . . . . . a_{8} - a_{9} + a_{10}$

⇒ $a_{0} - a_{1} + a_{2} - a_{3} + . . . . . . . . . . . . . - a_{9} + a_{10} = 2^{5}$ ...............(2)

(1) + (2)

$\Rightarrow$ 2 $(a_{0} - a_{2} + a_{4} . . . . . a_{10})$ = $2^{5}$

$a_{0} - a_{2} + a_{4} . . . . . a_{10}$ = $2^{4}$ = 16

To finf $a_{0}$ , put x = 0

$\Rightarrow$ $(- 1)^{5} = a_{0} = - 1$

$\Rightarrow$ -1 + $a_{2} + a_{4} . . . . . . a_{10}$ = 16

$a_{2} + a_{4} + a_{6} + a_{8} + a_{10}$ = 16 + 1

=17

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