Let $\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{20} x^{20}$. If $\left(a_1+a_3+a_5+\ldots .+a_{19}\right)-11 \mathrm{a}_2=121 \mathrm{k}$, then k is equal to $\qquad$ .

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Aicharya · Accepted Answer

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

Let $\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{20} x^{20}$. If $\left(a_1+a_3+a_5+\ldots …

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