数列 {an}\left\{ {a}_{n}\right\}{an} 的前 nnn 项和为 Sn,an+1=Snan(n∈N∗){S}_{n},{a}_{n + 1} = \frac{{S}_{n}}{{a}_{n}}\left( {n \in {\mathbf{N}}^{ * }}\right)Sn,an+1=anSn(n∈N∗) ,则 ∑i=15a2i−∑i=16a2i−1\mathop{\sum }\limits_{{i = 1}}^{5}{a}_{2i} - \mathop{\sum }\limits_{{i = 1}}^{6}{a}_{{2i} - 1}i=1∑5a2i−i=1∑6a2i−1 可以是（ ）A. 18B. 12C. 9D. 6