Question

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

• A. $a_0+a_2+...+a_{18}=a_1+a_3+...+a_{17}$
• B. $a_0+a_2+...+a_{18}$ is even
• C. $a_0+a_2+...+a_{18}$ is divisible by $9$
• D. $a_0+a_2+...+a_{18}$ is divisible by $3$ but not by $9$

Answer 1

The correct option is B $a_0+a_2+...+a_{18}$ is even

$(1+x+x^2{)}^9=a_0+a_{1}x+a_2{x}^2+...+a_{18}{x}^{18}$
Put $x=-1$ in the above equation, we have
$1=a_0-a_1+a_2-...+a_{18}\Rightarrow 1+a_1+a_3+...+a_{17}=a_0+a_2+...+a_{18}$

Now put $x=1$, we have
$3^9=a_0+a_1+a_2+...+a_{17}+a_{18}\Rightarrow 3^9+1=2(a_0+a_2+...+a_{18})\Rightarrow a_0+a_2+...+a_{18}=\frac{3^9+1}{2}=\frac{{27}^3+1^3}{2}=\frac{(27+1)({27}^2+1-27)}{2}=14({27}^2-26)$
which is an even number.