We are given that $(1+x)(2-x)^{2015}=a\_0+a\_1x+a\_2x^{2}+…+a_{2015}x^{2015}+a_{2016}x^{2016}$.

When $x=-1$, we have $0=a\_0-a\_1+a\_2+…-a_{2015}+a_{2016}$.

When $x=1$, we have $2=a\_0+a\_1+a\_2+…+a_{2015}+a_{2016}$.

When $x=0$, we have $2^{2015}=a\_0$.

By adding the equations obtained from $x=-1$ and $x=1$, and subtracting the equation obtained from $x=0$, we get:

$a\_2+a\_4+…+a_{2014}+a_{2016}=1-2^{2015}$.

Hence, the answer is $\boxed{1-2^{2015}}$.