Binomial Theorem

As we know that, coefficient of $x^{11}$ in the expansion of $(1+x)^{11}$ is ${}^{11}C_{11}$.Now, we have to find coefficient of $x^{11}$ in the expansion of $(1+\alpha x+\beta x^2) (1+x)^{11}$. When we choosing 1 from $(1+\alpha x+\beta x^2)$ then we need $x^{11}$ from $(1+x)^{11}$.So, coefficient $={}^{11}C_{11}=1$Similarly, we choosing $\alpha x$ from $(1+\alpha x+\beta x^2)$ then we need $x^{10}$ from $(1+x)^{11}=$ Coefficient $=\alpha \times {}^{11}C_{10}=11\alpha$ and when we choosing $\beta x^2$ from $(1+\alpha x+\beta x^2)$ then we need $x^9$ from $(1+x)=$ Coefficient $=\beta \times {}^{11}C_9=55 \cdot \beta$∴ Coefficient of $x^{11}$ in the expansion of $(1+\alpha x+\beta x^2)$$(1+x)^{11}=1+11\alpha+55\beta=144 \;\text{( given )}\;$Just, in similar way, we have to find coefficient of $x^{10}$ such that$11+55\alpha+165\beta=396$On solving Eqs. (i) and (ii), we get;$\alpha=-2 \;\text{and}\; \beta=3$$\therefore \alpha^2+\beta^2=(-2)^2+(3)^2=13$