Finding variance of a point estimator I'm having trouble understanding the solution for an example of finding the variance of a point estimator. Here is the question

I'm confused about the whole last line of the explanation. How do we go from $V(2Y)$ to $4V(Y)$ and then divide that by n?

Comment [see my Comment and the excellent Answer by @Clarinetist, (+1)]: 
Here is a simulation that may help you visualize the distributions of the two estimators and their relative dispersions. 
In my simulation
I take $\theta = 5$ and $n = 9.$ The simulated distributions are
based on a million such samples from $Unif(0, \theta).$
 m = 10^6;  th=5;  n = 9.
 x = runif(m*n, 0, th)
 DTA = matrix(x, nrow=m) # m x n matrix; each row a sample
 a = rowMeans(DTA)       # vector of m sample means
 w = apply(DTA, 1, max)  # vector on m sample maximums
 est.1 = 2*a
 est.2 = ((n+1)/n)* w
 mean(est.1);  sd(est.1)
 ## 4.999734   # est of E(est.1) = 5 (unbiased)
 ## 0.962236   # compare with 0.9622504 (from example)
 mean(est.2);  sd(est.2)
 ## 4.999626   # est of E(est.2) = 5 (unbiased)
 ## 0.502947   # markedly smaller than for est.1

The figure below clearly shows that the second estimator
(based on the maximum) has the smaller variance.

