NIMCET 2021 Question Paper with solutions

We have $\log \left(1-\mathit{x}+{\mathit{x}}^2\right)={\mathit{a}}_0+{\mathit{a}}_1\mathit{x}+{\mathit{a}}_2{\mathit{x}}^2$ $+{\mathit{a}}_3{\mathit{x}}^3+......$
Now $\log \left(1-\mathit{x}+{\mathit{x}}^2\right)=\log \left(1+\mathrm{\omega }\mathit{x}\right)\left(1+{\mathrm{\omega }}^2\mathit{x}\right)$
$=\log \left(1+\mathrm{\omega }\mathit{x}\right)+\log \left(1+{\mathrm{\omega }}^2\mathit{x}\right)......$(1)
We know that
$\log \left(1+\mathit{x}\right)=\mathit{x}-\frac{{\mathit{x}}^2}{2}+\frac{{\mathit{x}}^3}{3}-\frac{{\mathit{x}}^4}{4}$ $+........{\left(-1\right)}^{\mathit{n}+1}\frac{{\mathit{x}}^{\mathit{n}}}{\mathit{n}}$
So Eq. (1) can be written as
$\log \left(1+\mathrm{\omega }\mathit{x}\right)+\log \left(1+{\mathrm{\omega }}^2\mathit{x}\right)$
$\Rightarrow \left(\mathrm{\omega }\mathit{x}-\frac{{\left(\mathrm{\omega }\mathit{x}\right)}^2}{2}+\frac{{\left(\mathrm{\omega }\mathit{x}\right)}^3}{3}-.......\right)$ $+\left({\mathrm{\omega }}^2\mathit{x}-\frac{{\left({\mathrm{\omega }}^2\mathit{x}\right)}^2}{2}+\frac{{\left({\mathrm{\omega }}^2\mathit{x}\right)}^3}{3}-........\right)$
General value can be written as in the form
$\frac{{\left(-1\right)}^{\mathit{n}+1}{\mathrm{\omega }}^{\mathit{n}}}{\mathit{n}}+\frac{{\left(-1\right)}^{\mathit{n}+1}{\mathrm{\omega }}^{2\mathit{n}}}{\mathit{n}}$
$\Rightarrow \frac{{\left(-1\right)}^{\mathit{n}+1}}{\mathit{n}}\left({\mathrm{\omega }}^{\mathit{n}}+{\mathrm{\omega }}^{2\mathit{n}}\right).........\left(2\right)$
We have to calculate
${\mathit{a}}_3+{\mathit{a}}_6+{\mathit{a}}_9+......$
So from (2)
${\mathit{a}}_3=\frac{{\left(-1\right)}^{3+1}}{3}\left({\mathrm{\omega }}^3+{\mathrm{\omega }}^6\right)=\frac{+2}{3}$
${\mathit{a}}_6=\frac{{\left(-1\right)}^{6+1}}{6}\left({\mathrm{\omega }}^6+{\mathrm{\omega }}^{12}\right)=-\frac{2}{6}$
${\mathit{a}}_9=\frac{{\left(-1\right)}^{9+1}}{9}\left({\mathrm{\omega }}^9+{\mathrm{\omega }}^{18}\right)=\frac{+2}{9}$
⇒ ${\mathit{a}}_3+{\mathit{a}}_6+{\mathit{a}}_9+........$
⇒ $\frac{2}{3}-\frac{2}{6}+\frac{2}{9}-......$
⇒ $\frac{+2}{3}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}.......\right)$ $\Rightarrow \frac{2}{3}\log 2$