Let 1 - x - 2x2 20 - a0 - a1x -a2x2 - - a40x 40- Then- a1 - a3 -a5 - - - a37 is equal to-

Put x = 1, –1and subtract 4^20 – 2^20 = (a_0 + a_1 + ⋯ + a_40) – (a_0 – a_1 + ⋯) ⇒ 4^20 – 2^20=2(a_1+a_3+⋯ +a_39) ⇒ a_1+a_3+⋯ +a_37=2^39 – 2^19 a_39 a_39 = coef ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Representation of a Complex Number

Let 1 + x + 2...

Question

Let $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40} .$ Then, $a_{1} + a_{3} + a_{5} + \dots + a_{37}$ is equal to:

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

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

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2^{20} (2^{20} - 21)

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

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Solution

The correct option is D $2^{19} (2^{20} - 21)$
Put $x = 1, - 1$ and subtract
$4^{20} - 2^{20} = (a_{0} + a_{1} + \dots + a_{40}) - (a_{0} - a_{1} + \dots)$
$\Rightarrow 4^{20} - 2^{20} = 2 (a_{1} + a_{3} + \dots + a_{39})$
$\Rightarrow a_{1} + a_{3} + \dots + a_{37} = 2^{39} - 2^{19} - a_{39}$
$a_{39} =$ coeff of $x^{39}$ in $(1 + x + 2 x^{2})^{20} =^{20} C_{1} 2^{19}$
$\Rightarrow a_{1} + a_{3} + \dots + a_{37} = 2^{39} - 2^{19} - 20 (2^{19})$
$= 2^{39} - 21 (2^{19}) = 2^{19} (2^{20} - 21)$

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Similar questions

If $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} + a_{2} x^{2} + . . . + a_{40} x^{40}, t h e n a_{1} + a_{3} + a_{5} + . . . + a_{37} e q u a l s$ :

Q. If

(1 + x - 2 x^{2})^{8} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{16} x^{16}

then the sum

a_{1} + a_{3} + a_{5} + . . . . + a_{15}

is equal to

Q. If

(1 + x)^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{10} x^{10}

, then

(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} +

(a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

is equal to

Q. If

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

then

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

equals -

Q. Let

(1 + x + x^{2})^{2014} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . + a_{4028} x^{4028}

and let

A = a_{0} - a_{3} + a_{6} - . . . . . + a_{4026}

B = a_{1} - a_{2} + a_{7} - . . . . . + a_{4027}

C = a_{2} - a_{5} + a_{8} - . . . . . + a_{4028}

Then

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Let (1+x+2x2)20=a0+a1x+a2x2+⋯+a40x40. Then, a1+a3+a5+⋯+a37 is equal to:

The correct option is D 219(220−21)Put x=1,–1and subtract 420–220=(a0+a1+⋯+a40)–(a0–a1+⋯) ⇒420–220=2(a1+a3+⋯+a39) ⇒a1+a3+⋯+a37=239–219−a39 a39= coeff of x39 in (1+x+2x2)20=20C1219 ⇒a1+a3+⋯+a37=239–219−20(219) =239−21(219)=219(220−21)

Let $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40} .$ Then, $a_{1} + a_{3} + a_{5} + \dots + a_{37}$ is equal to: