Question

lf $\log (1-x+x^2)=a_{1}x+a_2{x}^2+a_3{x}^3+\dots$. then
$a_3+a_6+a_9+\cdots \cdot$

• A. $\log 2$
• B. $\frac{2}{3}\log 2$
• C. $\frac{1}{3}\log 2$
• D. $2\log 2$

Answer 1

Answer: (B)

$\frac{2}{3}\log 2$

Let
$x=w$ where $w$ is the cube root of unity.
Hence we know that $1+w+w^2=0$ and $w^3=1$.
And $w^{3n+k}=w^{3n}.w^k=w^k$ where both $n,kϵR$.
Therefore
$log(1-w+w^2)=a_{1}w+a_2{w}^2+a_3+a_{4}w+a_5{w}^2+a_6...$
Let $x=w^2$
$log(1-w^2+w)=a_1{w}^2+a_{2}w+a_3+a_4{w}^2+a_{5}w+a_6...$
Let $x=1$, we get
$log\left(1\right)=a_1+a_2+a_3+a_4...$
Adding i ii and iii, we get
$log(1-w^2+w)+log(1-w+w^2)+log\left(1\right)=a_1(1+w+w^2)+a_2(1+w+w^2)+3a_3+a_4(1+w+w^2)...$
$log(1-w^2+w)+log(1-w+w^2)+log\left(1\right)=3[a_3+a_6+a_9.....]$
$log\left(\right(1+w)-w^2)+log\left(\right(1+w^2)-w)+0=3[a_3+a_6+a_9.....]$
$log(-w^2-w^2)+log(-w-w)=3[a_3+a_6+a_9.....]$
$log(-2w^2)+log(-2w)=3[a_3+a_6+a_9.....]$
$log(-2w^2\times (-2w\left)\right)=3[a_3+a_6+a_9.....]$
$log\left(4w^3\right)=3[a_3+a_6+a_9.....]$
$log\left(4\right)=3[a_3+a_6+a_9.....]$
$\frac{2log\left(2\right)}{3}=[a_3+a_6+a_9.....]$