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NIMCET 2021 Question Paper with solutions

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Question : 13 of 120

Marks: +1, -0

If

\log (1-x+x^{2})=a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\dots

a_{3}+a_{6}+a_{9}+\dots

is equal to

$\log 2$
$\frac{2}{3} \log 2$
$\frac{1}{3} \log 2$
$2 \log 2$

Solution:

We have

\log (1-x+x^{2})=a_{0}+a_{1} x+a_{2} x^{2}

+a_{3} x^{3}+\dots

Now

\log (1-x+x^{2})=\log (1+\omega x)(1+\omega^{2} x)

=\log (1+\omega x)+\log (1+\omega^{2} x) \dots

(1)

We know that

\log (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}

+\dots .(-1)^{n+1} \frac{x^{n}}{n}

So Eq. (1) can be written as

\log (1+\omega x)+\log (1+\omega^{2} x)

\Rightarrow \left(\omega x-\frac{(\omega x)^{2}}{2}+\frac{(\omega x)^{3}}{3}-\dots\right)

+\left(\omega^{2} x-\frac{(\omega^{2} x)^{2}}{2}+\frac{(\omega^{2} x)^{3}}{3}-\dots\right)

General value can be written as in the form

\frac{(-1)^{n+1} \omega^{n}}{n}+\frac{(-1)^{n+1} \omega^{2n}}{n}

\Rightarrow \frac{(-1)^{n+1}}{n}(\omega^{n}+\omega^{2n})\dots .(2)

We have to calculate

a_{3}+a_{6}+a_{9}+\dots

So from (2)

a_{3}=\frac{(-1)^{3+1}}{3}(\omega^{3}+\omega^{6})=\frac{+2}{3}

a_{6}=\frac{(-1)^{6+1}}{6}(\omega^{6}+\omega^{12})=-\frac{2}{6}

a_{9}=\frac{(-1)^{9+1}}{9}(\omega^{9}+\omega^{18})=\frac{+2}{9}

⇒

a_{3}+a_{6}+a_{9}+\dots .

⇒

\frac{2}{3}-\frac{2}{6}+\frac{2}{9}-\dots

⇒

\frac{+2}{3}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\dots .\right)

\Rightarrow \frac{2}{3} \log 2

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