Question

If (1+x+x2)n = a0 +a1x +a2x2 + a3x3+ .......+ a2nx2n,show thata0+a3+a6 + .......... = 3n-1

Answer 1

Given :

(1+x+x²)ⁿ   =  a₀ +a₁x +a₂x₂+ a₃x³+ .......+ a₂ₙx²ⁿ

To Find : ,show that  a₀ +  a₃  +.....=3ⁿ⁻¹

Solution:

(1+x+x²)ⁿ   =  a₀ +a₁x +a₂x₂+ a₃x³+ .......+ a₂ₙx²ⁿ

x = 1

=> 3ⁿ = a₀ +a₁  +a₂ + a₃ + .......+ a₂ₙ

=> 3ⁿ = (a₀ +  a₃  +.....)  + (a₁ +.......)  + (a₂ +.....)

x =ω

=> (1+ω+ω²)ⁿ =  a₀+a₁ω  +a₂ω²+ a₃ω³ + .......+ a₂ₙω²ⁿ

ω³  = 1  , ω⁴ = ω  and so on

1+ω+ω²  = 0

=> 0 = (a₀ +  a₃  +.....)  + (a₁ +.......)ω  + (a₂ +.....)ω²

x =ω²

=> (1+ω²+ω⁴)ⁿ =  a₀+a₁ω²  +a₂ω⁴+ a₃ω⁶ + .......+ a₂ₙω²ⁿ

ω³  = 1  , ω⁴ = ω  and so on

1+ω²+ω⁴ = 1+ω² +ω = 0

=> 0 = (a₀ +  a₃  +.....)  + (a₁ +.......)ω²  + (a₂ +.....)ω

Adding all 3

=> 3ⁿ = (a₀ +  a₃  +.....) (1 + 1 + 1) + (a₁ +.......)(1+ω+ω²)  +  (a₂ +.....)(1 + ω² +ω)

=> 3ⁿ = (a₀ +  a₃  +.....) (3) + (a₁ +.......)(3)  +  (a₂ +.....)(0)

=> 3ⁿ = (a₀ +  a₃  +.....) 3

=> 3ⁿ/3 = a₀ +  a₃  +.....

=> a₀ +  a₃  +..... = 3ⁿ⁻¹

QED

Hence Proved

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\sf{ {(1 + x + {x}^{2}) }^{n} = a_0+a_1x+a_2 {x}^{2} +a_3 {x}^{3} ...+a_{2n} {x}^{2n} }$

TO PROVE

$\displaystyle \sf{ a_0+a_3+a_6 + ... = \frac{ {3}^{n} - 1}{2} }$

PROOF

It is given that

$\sf{ {(1 + x + {x}^{2}) }^{n} = a_0+a_1x+a_2 {x}^{2} +a_3 {x}^{3} ...+a_{2n} {x}^{2n} }...(1)$

Putting x = 1 in Equation (1) we get

$\sf{ {(1 + 1 + {1}^{2}) }^{n} = a_0+a_1+a_2 +a_3 ...+a_{2n} }$

$\implies\sf{ a_0+a_1+a_2 +a_3 ...+a_{2n} = {3}^{n} } \: \: \: .....(2)$

Putting x = - 1 in Equation (1) we get

$\sf{ {(1 - 1 + {1}^{2}) }^{n} = a_0 - a_1+a_2 - a_3 ...+a_{2n} }$

$\implies\sf{ a_0 - a_1+a_2 - a_3 ...+a_{2n} = 1 } \: \: \: .....(3)$

Equation (2) - Equation (3) gives

$\displaystyle \sf{ 2( a_0+a_3+a_6 +.. ...) = {3}^{n} - 1}$

$\implies \displaystyle \sf{ a_0+a_3+a_6 + ... = \frac{ {3}^{n} - 1}{2} }$

Hence proved

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