9．已知数列{an}\{a_{n}\}{an}满足2na1+2n−1a2+⋯+22an−1+2an=2n−n2−1{2}^{n}{a}_{1}+{2}^{n-1}{a}_{2}+\cdots +{2}^{2}{a}_{n-1}+2{a}_{n}={2}^{n}-\frac{n}{2}-12na1+2n−1a2+⋯+22an−1+2an=2n−2n−1，若cn=1an+an+1{c}_{n}=\frac{1}{\sqrt{{a}_{n}}+\sqrt{{a}_{n+1}}}cn=an+an+11，则数列{cn}\{c_{n}\}{cn}的前nnn项和Tn=T_{n}=Tn=___2(n+1−1)2(\sqrt{n+1}-1)2(n+1−1)___．