已知数列 {an}\left\{ {a}_{n}\right\}{an} 的前 nnn 项和为 Sn,a1=1,a2=2,anan+2=an+12+n(n∈N∗){S}_{n},{a}_{1} = 1,{a}_{2} = 2,{a}_{n}{a}_{n + 2} = {a}_{n + 1}^{2} + n\left( {n \in {\mathrm{N}}^{ * }}\right)Sn,a1=1,a2=2,anan+2=an+12+n(n∈N∗) ,则下列结论不正确的是 ( )A. {an+1an}\left\{ \frac{{a}_{n + 1}}{{a}_{n}}\right\}{anan+1} 是递增数列B. {an+12−an2}\left\{ {{a}_{n + 1}^{2} - {a}_{n}^{2}}\right\}{an+12−an2} 是递增数列C. S10<1023{S}_{10} < {1023}S10<1023D. an+1an<3\frac{{a}_{n + 1}}{{a}_{n}} < 3anan+1<3