Question

If $(1+x+x^2{)}^n=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+...+a_{2n}{x}^{2n}$ and $\begin{array}{ccc}A=a_0+a_3+a_6+...+a_{3k}+& ...& \\ B=a_1+a_4+a_7+...+a_{3k+1}+& ...& \\ C=a_2+a_5+a_8+...+a_{3k+2}+& ...& \end{array}\}\forall k=0,1,...n$ then which of following is/are true?

• A. $A+B+C=3^n$
• B. $A=B=C$
• C. $A=B=C=3^{n-1}$
• D. $A=B=C=3^{n-2}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $A+B+C=3^n$
B $A=B=C$
C $A=B=C=3^{n-1}$

${(1+x+x^2)}^n=a_0+a_{1}x+a_1{x}^2+...+a_{2n}{x}^{2n}$
$\therefore {(1+x+x^2)}^n=(a_0+a_3{x}^3+a_0{x}^6+...)$
$+x(a_1+a_4{x}^3+...)+x^2(a_2+a_3{x}^3+a_8{x}^6+...)$...(i)
Putting $x=1,\omega ,{\omega }^2$
$3^n=(a_0+a_3+a_6+...)+(a_1+a_4+a_7+...)+(a_2+a_5+a_5+...)$
$3^n=A+B+C$
Again $0=A+B\omega +C{\omega }^2...(\ast \ast )$
$0=A+B\omega +C\omega ...(\ast \ast \ast )$
By adding (*), (**)and (***) we get
$A=3^{n-1}=B=C$