12.设(2x+1)5=a0+a1x+⋅⋅⋅+a5x5{(2x+1)}^{5}={a}_{0}+{a}_{1}x+\cdot \cdot \cdot +{a}_{5}{x}^{5}(2x+1)5=a0+a1x+⋅⋅⋅+a5x5,则a1+a2+…+a5=(a_{1}+a_{2}+\ldots +a_{5}=(a1+a2+…+a5=( )))