If 1-x-x2n-a0-a1 x-a2 x2-a2 n x2 n- and E1-a0-a3-a6-E2-a1-a4-a7-E3-a2-a5-a8- thenA- E2-3n-1B- E2-3nC- E3-3n-1D- E3-3n

The correct options are A E_2=3^n 1 C E_3=3^n 1(1+x+x^2)^n=a_0+a_1x+a_2x^2+a_3x^3+...+a_2nx^2n ⇒ (1+x+x^2)^n=(a_0+a_3x^3+a_6x^6+...) +x(a_1+a_4x^3+a_7x^6+...)+x ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

If 1+x+x2n=a0...

Question

If $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}$ , and $E_{1} = a_{0} + a_{3} + a_{6} + . . .$ ,
$E_{2} = a_{1} + a_{4} + a_{7} + . . .$ ,
$E_{3} = a_{2} + a_{5} + a_{8} + . . .$ , then

E_{2} = 3^{n - 1}

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E_{2} = 3^{n}

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E_{3} = 3^{n - 1}

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E_{3} = 3^{n}

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Solution

The correct options are
A $E_{2} = 3^{n - 1}$
C $E_{3} = 3^{n - 1}$
$(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . + a_{2 n} x^{2 n}$
$\Rightarrow (1 + x + x^{2})^{n} = (a_{0} + a_{3} x^{3} + a_{6} x^{6} + . . .)$
$+ x (a_{1} + a_{4} x^{3} + a_{7} x^{6} + . . .) + x^{2} (a_{2} + a_{5} x^{3} + a_{8} x^{6} + . . .)$

Putting $x = 1, ω, ω^{2}$ respectively, we get
$3^{n} = E_{1} + E_{2} + E_{3}$ ....... $(1)$
$0 = E_{1} + ω E_{2} + ω^{2} E_{3}$ ...... $(2)$
$0 = E_{1} + ω^{2} E_{2} + ω E_{3}$ ...... $(3)$

$(1) + (2) + (3) \Rightarrow$
$3 E_{1} = 3^{n} \Rightarrow E_{1} = 3^{n - 1}$

$(1) + (2) \times ω^{2} + (3) \times ω \Rightarrow$
$3 E_{2} = 3^{n} \Rightarrow E_{2} = 3^{n - 1}$
$\Rightarrow E_{3} = 3^{n - 1}$

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Similar questions

Q. If

(1 + x + x^{2}) n = a 0 + a^{1} x + a^{2} x + a^{3} x^{3} + . . . . . . . a (2^{n}) x (2^{n})

then prove that

(a^{0} + a^{3} + a^{6} + . . . . . . . . .) = (a^{1} + a^{4} + a^{7}) = (a^{2} + a^{5} + a^{8}) = 3^{(} n - 1),

Q. If the expansion of

(1 + x + x^{2})^{n}

a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{r} x^{r} + . . . . . + a_{2 n} x^{2 n}

,
show that

a_{0} + a_{3} + a_{6} + . . . . . = a_{1} + a_{4} + a_{7} + . . . . . = a_{2} + a_{5} + a_{8} + . . . . = 3^{n - 1}

Q. If

a_{0}, a_{1}, a_{2}, . . . .

be the coefficients in the expansion of

{(1 + x + x^{2})}^{n}

in ascending powers of

x

, also if

E_{1} = a_{0} + a_{3} + a_{6} + . . .,

E_{2} = a_{1} + a_{4} + a_{7} . . . .

and

E_{3} = a_{2} + a_{5} + a_{8} + . . .;

then:

Q. If

{(1 + x + x^{2})}^{n} = 2 n \sum r = 0 a_{r} x^{r} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a^{2 n} x^{2 n}

and

P = a_{0} + a_{3} + a_{6} + . . .

Q = a_{1} + a_{4} + a_{7} + . . .

R = a_{2} + a_{5} + a_{8} + . . .

then the set of values of

P, Q, R

are respectively equals

Q. If

a_{0}, a_{1}, a_{2}, . . . . .

be the coefficients in the expansion of

(1 + x + x^{2})^{n}

in ascending powers of x, then prove that :

(a_{0} + a_{3} + a_{6} + . . .) = (a_{1} + a_{4} + a_{7} + . . . .) = = (a_{2} + a_{5} + a_{8} + . . .) = 3^{n - 1}

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If (1+x+x2)n=a0+a1x+a2x2+...+a2nx2n, and E1=a0+a3+a6+..., E2=a1+a4+a7+..., E3=a2+a5+a8+..., then

If $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{2 n} x^{2 n}$ , and $E_{1} = a_{0} + a_{3} + a_{6} + . . .$ ,
$E_{2} = a_{1} + a_{4} + a_{7} + . . .$ ,
$E_{3} = a_{2} + a_{5} + a_{8} + . . .$ , then