If $\left(1+x+x^2\right)^n=c_0+c_1 x+c_2 x^2+\ldots$, then the value of $c_0 c_1-c_1 c_2+c_2 c_3-\ldots$ is

Question

Aicharya · Accepted Answer

On replacing $x$ by $-\frac{1}{x}$, we get $ \begin{aligned} & \frac{\left(1-x+x^2\right)^n}{x^{2 n}} \\ & =\frac{C_0 x^{2 n}-C_1 x^{2 n-1}+C_2 x^{2 n-2}-C_3 x^{2 n-3}+\ldots}{x^{2 n}} \\ & \Rightarrow\left(1-x+x^2\right)^n=C_0 x^{2 n}-C_1 x^{2 n-1}+C_2 x^{2 n-2} \end{aligned} $

$ \begin{aligned} & \because C_0 C_1-C_1 C_2+C_2 C_3-\ldots=\text { Coefficient of } x^{2 n+1} \text { in } \\ & \left(C_0 x^{2 n}-C_1 x^{2 n-1}+C_2 x^{2 n-2}-C_3 x^{2 n-3}+\ldots\right) \\ & \times\left(C_0+C_1 x+C_2 x^2+C_3 x^3+\ldots\right) \\ & =\text { Coefficient of } x^{2 n+1} \text { in }\left(1+x+x^2\right)^n\left(1-x+x^2\right)^n \\ & =\text { Coefficient of } x^{2 n+1} \text { in }\left(\left(1+x^2\right)^2-x^2\right)^n \\ & =\text { Coefficient of } x^{2 n+1} \text { in }\left(1+x^4+x^2\right)^n=0 \end{aligned} $ $\because$ Coefficient of $x^k$ in the expansion of $\left(1+x^2+x^4\right)^n$ exists if $k$ is even, otherwise it will be zero and $(2 n+1)$ is odd. Hence, option (b) is correct

If $\left(1+x+x^2\right)^n=c_0+c_1 x+c_2 x^2+\ldots$, then the value of $c_0 c_1-c_1 c_2+c_2 c_3-\ldots$ is

Solution