Let 1-x-x410-a0-a1 x-a2 x2-a40 x40- then the correct options is-are A- a0-a2-a4-a40-310-1-2B- a 0- a 40-1C- a 1- a 39D- a39-0

The correct options are A a_0+a_2+a_4+.....+a_40=3^10+1/2 B a_0=a_40=1 D a_39=0(1 x+x^4)^10=a_0+a_1 x+a_2x^2+.....+a_40x^40 put x=1 ⇒ nbsp;a_0+a_1 +a_2+... ...

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Byju's Answer

Standard XII

Mathematics

Multinomial Expansion

Let 1-x+x410=...

Question

Let ${(1 - x + x^{4})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{40} x^{40}$ , then the correct option(s) is/are

a_{0} + a_{2} + a_{4} + . . . . . + a_{40} = \frac{3^{10} + 1}{2}

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a_{0} = a_{40} = 1

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a_{1} = a_{39}

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a_{39} = 0

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Solution

The correct options are
A $a_{0} + a_{2} + a_{4} + . . . . . + a_{40} = \frac{3^{10} + 1}{2}$
B $a_{0} = a_{40} = 1$
D $a_{39} = 0$
${(1 - x + x^{4})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{40} x^{40} put x = 1 \Rightarrow a_{0} + a_{1} + a_{2} + . . . . . + a_{40} = 1 put x = - 1 \Rightarrow a_{0} - a_{1} + a_{2} - . . . . . + a_{40} = 3^{10} adding both a_{0} + a_{2} + a_{4} + . . . . . + a_{40} = \frac{3^{10} + 1}{2} In original equaiton put x = 0 a_{0} = 1 replace x \to \frac{1}{x} \Rightarrow {(1 - \frac{1}{x} + \frac{1}{x^{4}})}^{10} = a_{0} + \frac{a_{1}}{x} + \frac{a_{2}}{x^{2}} + . . . . . + \frac{a_{40}}{x^{40}} \Rightarrow {(x^{4} - x^{3} + 1)}^{4} = a_{0} x^{40} + a_{1} x^{39} + a_{2} x^{38} + . . . . + a_{40} Putting x = 0, we get a_{40} = 1 differentiate and put x = 0 a_{39} = 0$

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Similar questions

$(1 + x + 2 x^{2})^{20}$ = $a_{0} + a_{1} x + a_{2} x^{2}$ ................... $a_{40} x^{40}$ , If $a_{0} + a_{1} + a_{2} + a_{3} . . . . . . . . . . . . . . . . . . a_{40} = 2^{a}$

$a_{0} - a_{1} + a_{2} - a_{3} . . . . . . . . . . . . . . . . . . . . . . a_{40} = 2^{b}$ . Find the value of a + b

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Q. Let

{(1 - x + x^{4})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40} .

Then the correct option(s) is (are)

Q. If

{(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}

If $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} . . . . . a_{40} x^{40}$ , then $a_{0} + a_{2} + a_{4} . . . . a_{38}$ =

Q. Let

(1 + x + x^{2} + x^{3} + x^{4})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{40} x^{40}

. Then:

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Let (1−x+x4)10=a0+a1x+a2x2+.....+a40x40, then the correct option(s) is/are

Let ${(1 - x + x^{4})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{40} x^{40}$ , then the correct option(s) is/are