The random variable X follows binomial distribution B (n, p), for which the difference of the mean and the variance is 1. If$2P(X=2)=3P(x=1)$, then $n^{2}P( X > 1)$ is equal to

$$\begin{gathered} \begin{array}{l}2P(X=2)=3P(x=1) \\ \Rightarrow \text{ 2. }{\mathrm{n}}_{{\mathrm{c}}_2}{\mathrm{p}}^2\cdot (1-\mathrm{p}{)}^{\mathrm{n}-2}=3{\cdot }^{\mathrm{n}}{\mathrm{C}}_1\mathrm{p}(1-\mathrm{p}{)}^{\mathrm{n}-1}\\ \text{ 2. }\frac{\mathrm{n}(\mathrm{n}-1)}{2}\mathrm{p}=3.\mathrm{n}(1-\mathrm{p})\\ \Rightarrow (\mathrm{n}-1)\mathrm{p}=3-3\mathrm{p}\Rightarrow (\mathrm{n}+2)\mathrm{p}=3\Rightarrow \mathrm{p}=\frac{3}{\mathrm{n}+2}\\ \begin{array}{l}Mean-variance=\mathrm{np}-\mathrm{npq}=1\Rightarrow {\mathrm{np}}^2=1\Rightarrow {\mathrm{p}}^2=\frac{1}{\mathrm{n}}\\ \Rightarrow \frac{9}{(\mathrm{n}+2{)}^2}=\frac{1}{\mathrm{n}}\Rightarrow {\mathrm{n}}^2+4\mathrm{n}+4=9\mathrm{n}\\ \Rightarrow {\mathrm{n}}^2-5\mathrm{n}+4=0\\ \Rightarrow \mathrm{n}=4\\ \therefore \mathrm{p}=\frac{3}{6}=\frac{1}{2}\end{array}\end{array} \\ \begin{array}{l}{\mathrm{n}}^2\mathrm{p}(\mathrm{x}>1)=16(1-\mathrm{p}(\mathrm{x}=0)-\mathrm{p}(\mathrm{x}=1\left)\right)\\ =16\left(1-4_{{\mathrm{c}}_0}{\left(\frac{1}{2}\right)}^4-4_{{\mathrm{c}}_1}{\left(\frac{1}{2}\right)}^4\right)=11\end{array} \end{gathered}$$