If ${(2{x^2} - x - 1)^5} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{10}}{x^{10}},\;{\rm{then\;}}{a_2} + {a_4} + {a_6} + {a_8} + {a_{10}}$ is equal to

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

{(2{x^2} - x - 1)^5} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{10}}{x^{10}},\;\mathrm{then\;}{a_2} + {a_4} + {a_6} + {a_8} + {a_{10}}

is equal to

View answer and explanation

Correct option: A

Option A (15) is correct because by substituting x = 1 and x = -1 into the given equation and adding the resulting equations, we can isolate and find the value of

{a_2} + {a_4} + {a_6} + {a_8} + {a_{10}}

More Questions from BINOMIAL THEOREM

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\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

equals If

a

and

d

are two complex numbers, then the sum to

\left( {n + 1} \right)

terms of the following series

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

is The number of different terms in the expansion of

(1-x)^{201}\left(1+x+x^2\right)^{200}

is The sum of the coefficient of the binomial expansion of

\left(\frac{1}{x}+2 x\right)^n

is equal to 6561 . The constant term in the expansion is The coefficient of

{x^5}

in the expansion of

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

is If

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

, then

n

equals