Mathematics - Binomial Theorem Question with Solution | TestHub

${\left(1+x+x^2\right)}^{10}=a_0+a_{1}x+a_2{x}^2+\dots .+a_{20}{x}^{20}$ $\therefore 3^{10}={\mathrm{a}}_0+{\mathrm{a}}_1+{\mathrm{a}}_2+\dots .+{\mathrm{a}}_{20}$ ...(i) $1=a_0-a_1+a_2\dots ..+a_{20}$ ...(ii) $\text{ (i) - (ii) }\Rightarrow a_1+a_3+\dots .+a_{19}=\frac{3^{10}-1}{2}=29524$ $\begin{array}{rl}\text{ Also }\{& 1+\mathrm{x}(1+\mathrm{x}){\}}^{10}=1\\ & +{}^{10}{\mathrm{C}}_1\mathrm{x}(1+\mathrm{x})+{}^{10}{\mathrm{C}}_2{\mathrm{x}}^2(1+\mathrm{x}{)}^2+\dots .\end{array}$ $\therefore {\mathrm{a}}_2={}^{10}{\mathrm{C}}_1+{}^{10}{\mathrm{C}}_2=55$ $\therefore \frac{\left({\mathrm{a}}_1+{\mathrm{a}}_3+\dots +{\mathrm{a}}_{19}\right)-11{\mathrm{a}}_2}{121}=239$