imc

The sum of any four consecutive powers of 3 is divisible by $3^0+3^1+3^2+3^3=40$ and hence is divisible by 8. Therefore $(3^2+3^3+3^4+3^5)+\cdots +(3^{2006}+3^{2007}+3^{2008}+3^{2009})$ is divisible by 8. So the required remainder is $3^0+3^1=\boxed{4}$. The answer is $(\mathrm{D})$.