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Greatest Binomial Coefficients

If 1+x+2x22...

Question

If ${(1 + x + 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{40}$ , then $a_{1} + a_{3} + a_{5} + . . . . . . + a_{39}$ equals

A

2^{19} (2^{20} - 1)

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B

2^{20} (2^{19} - 19)

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C

2^{19} (2^{20} + 21)

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D

None of these

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Solution

The correct option is A $2^{19} (2^{20} - 1)$
$(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . . . . + a_{40} x^{40}$
$x = 1$
$(1 + 1 + 2)^{20} = a_{0} + a_{1} + a_{2} + a_{3} + . . . . + a_{40}$ ________ (1)
$x = - 1$
$(1 - 1 + 2)^{20} = a_{0} - a_{1} + a_{2} - a_{3} + . . . . . + a_{40}$ _________(2)
(1) _ (2)
$4^{20} - 2^{20} = 2 (a_{1} + a_{3} + . . . . . . . . + a_{39})$
$2^{20} (2^{20} - 1) = 2 (a_{1} + a_{3} + . . . . . . + a_{39})$
$a_{1} + a_{3} + . . . . . . + a_{39} = 2^{19} (2^{20} - 1)$

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Similar questions

{(1 + x + 2 x^{2})}^{20} = 0 + a_{1} x + a_{2} x^{2} + . . . . . . + a_{40} x^{40}

a_{1} + a_{3} + a_{5} + . . . . . . + a_{37}

{(1 + x + 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{39} x^{39} + a_{40} x^{40}

then find the value of

a_{0} + a_{1} + a_{2} + . . . . . . + a_{38}

(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . a_{40} x^{40}

. Find the values of

a_{1} + a_{3} + a_{5} + . . . . a_{39}

{(1 + x - 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}

and the value of

a_{1} + a_{3} + a_{5} + \dots + a_{39} = - 2^{k}

k =

{(1 + x - 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}

and the value of

a_{1} + a_{3} + a_{5} + \dots + a_{39} = - 2^{k}

k =

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