If `(1+x+x^(2))^(n) = a_(0) + a_(1)x+ a_(2)x^(2) + "……" a_(2n)x^(2n)`, find the value of `a_(0) + a_(3) + a_(6) + "……", n in N`.

Options

A. `a_(1) + a_(4) +a_(7) + …`
B. `a_(2) +a_(5) +a_(8) +…`
C. `3^(n-1)`
D. all of these

Answer

d

Explanation

We have,
`(1 + x + x^(2))^(n) = (a_(0) + a_(3) x^(3)+ a_(6) x^(6) +…)
+ x (a_(1) +a_(4) x^(3) +…) + x^(2)(a_(2) + a_(5) x^(3) +)`
Putting x=1 and `omega` and `omega^(2) ` respectively, we get
`3^(n) = (a_(0) + a_(3) +a_(6) +...)+ (a _(1) + a_(4) + ....) + (a_(2) + a_(5)+...)` ....(i)
`0 = (a_(0) + a_(3) +a_(6) +...)+omega(a _(1) + a_(4) + ....) + omega^(2)(a_(2) + a_(5)+...)` ....(ii)
`0 = (a_(0) + a_(3) +a_(6) +...)+omega^(2)(a _(1) + a_(4) + ....) + omega(a_(2) + a_(5)+...)` .....(iii)
Adding these , we get
`3^(n) = (a_(0) + a_(3) +a_(6) +...)rArr a _(1) + a_(4) + ...3^(n-1)`
Multiplying (ii) and (iii) by `omega^(2)` and `omega` respectively and adding ltbRgt (i). (ii) and (iii), we get
`a_(1) + a_(4) + a_(7) + ..= 3^(n-1)`
Similarly, we have
`a_(2) + a_(5) + a_(8) + ...= 3^(n-1)` .