若锐角 θ\thetaθ 满足 tan2θ=23cosθ\tan {2\theta } = 2\sqrt{3}\cos \thetatan2θ=23cosθ ,数列 {an}\left\{ {a}_{n}\right\}{an} 的前 nnn 项和为 Sn,a1=1,nan+1=(109+cos4θ)(n+1)an{S}_{n},{a}_{1} = 1,n{a}_{n + 1} = \left( {\frac{10}{9} + \cos {4\theta }}\right) \left( {n + 1}\right) {a}_{n}Sn,a1=1,nan+1=(910+cos4θ)(n+1)an ,则使得 Sn+3n2⋅3n<5625{S}_{n} + \frac{3n}{2 \cdot {3}^{n}} < \frac{56}{25}Sn+2⋅3n3n<2556 成立的 nnn 的最大值为( )A. 2B. 3C. 4D. 5