(1+x+x2)10=a0+a1x+a2x2+….+a20x20 ∴310=a0+a1+a2+….+a20 ...(i) 1=a0−a1+a2…..+a20 ...(ii) (i) - (ii) ⇒a1+a3+….+a19=2310−1=29524 $\begin{aligned}
\text { Also }{ & 1+\mathrm{x}(1+\mathrm{x})}^{10}=1 \
& +{ }^{10} \mathrm{C}_1 \mathrm{x}(1+\mathrm{x})+{ }^{10} \mathrm{C}_2 \mathrm{x}^2(1+\mathrm{x})^2+\ldots .
\end{aligned}\therefore \mathrm{a}_2={ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_2=55\therefore \frac{\left(\mathrm{a}_1+\mathrm{a}3+\ldots+\mathrm{a}{19}\right)-11 \mathrm{a}_2}{121}=239$