设随机变量$X\sim B(n,p)$，其中$n\ge 4$且$n\in {{\mathbf{N}}_{+}}$，$p\in (0,1)$，若$E(3X)=E(2n-X)$，$3P(X=3)=4P(X=4)$，则$D(pX+n)=$                 ．

$\because X\sim B(n,p)$，

$\therefore E(X)=np$，$E(3X)=3np$，$E(2n-X)=2n-np$，

由$E(3X)=E(2n-X)$，得$4np=2n$，

$\therefore p=\dfrac{1}{2}$，

$P(X=k)=\operatorname{\rm {C}}_{n}^{k}{{\left( \dfrac{1}{2} \right)}^{n}}(k=0,1,\cdot \cdot \cdot ,n)$，

由$3P(X=3)=4P(X=4)$，得$3\operatorname{\rm {C}}_{n}^{3}{{\left( \dfrac{1}{2} \right)}^{n}}=4\operatorname{\rm {C}}_{n}^{4}{{\left( \dfrac{1}{2} \right)}^{n}}$，即$3\operatorname{\rm {C}}_{n}^{3}=4\operatorname{\rm {C}}_{n}^{4}$，解得$n=6$，

$\therefore D(X)=np(1-p)=\dfrac{3}{2}$，$D(pX+n)={{p}^{2}}D(X)=\dfrac{1}{4}\times \dfrac{3}{2}=\dfrac{3}{8}$．

故答案为：$\dfrac{3}{8}$