How can I find the order of growth?

Question

$$\sum_{i=1}^n i2^{n+1}$$

Is it $\Theta(n^2)$ as the $2^{n+1}$ is a constant so it will drop out and it will be only $i$ which is $\cfrac{n(n+1)}{2}$?

Answer 1

As has been pointed out in the comments, $$f(n)=\sum_{i=1}^ni2^{n+1}=n^22^n+n2^n$$ so $f(n)$ is asymptotic to $n^22^n$.