Let (1+x+x^2)^9=a_0+a_1 x+a_2 x^2 +ldots+a_{1 | vedclass

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(B) Given $(1+x+x^2)^9 = a_0 + a_1 x + a_2 x^2 + \ldots + a_{18} x^{18}$. Put $x = 1$, we get $(1+1+1)^9 = a_0 + a_1 + a_2 + \ldots + a_{18} \Rightarr

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$a_0+a_2+\ldots+a_{18}$ is even

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(B) Given $(1+x+x^2)^9 = a_0 + a_1 x + a_2 x^2 + \ldots + a_{18} x^{18}$. Put $x = 1$, we get $(1+1+1)^9 = a_0 + a_1 + a_2 + \ldots + a_{18} \Rightarrow 3^9 = a_0 + a_1 + a_2 + \ldots + a_{18} \quad (i)$. Put $x = -1$, we get $(1-1+1)^9 = a_0 - a_1 + a_2 - a_3 + \ldots + a_{18} \Rightarrow 1 = a_0 - a_1 + a_2 - a_3 + \ldots + a_{18} \quad (ii)$. Adding $(i)$ and $(ii)$, we get $3^9 + 1 = 2(a_0 + a_2 + a_4 + \ldots + a_{18})$. Therefore, $a_0 + a_2 + a_4 + \ldots + a_{18} = \frac{3^9 + 1}{2} = \frac{19683 + 1}{2} = \frac{19684}{2} = 9842$. Since $9842$ is an even number, option $B$ is correct.

Let $(1+x+x^2)^9=a_0+a_1 x+a_2 x^2 +\ldots+a_{18} x^{18}$. Then

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