已知 {an}\left\{ {a}_{n}\right\}{an} 为公比大于 0 的等比数列，其前 nnn 项和为 Sn{S}_{n}Sn ，且 a1=1,S2=a3−1{a}_{1} = 1,{S}_{2} = {a}_{3} - 1a1=1,S2=a3−1 .（1）求 {an}\left\{ {a}_{n}\right\}{an} 的通项公式及 Sn{S}_{n}Sn ；（2）设数列 {bn}\left\{ {b}_{n}\right\}{bn} 满足 bn={k,n=akbn−1+2k,ak<n<ak+1{b}_{n} = \left\{ \begin{array}{l} k,n = {a}_{k} \\ {b}_{n - 1} + {2k},{a}_{k} < n < {a}_{k + 1} \end{array}\right.bn={k,n=akbn−1+2k,ak<n<ak+1 ，其中 k∈N∗k \in {\mathbf{N}}^{ * }k∈N∗ .(i) 当 n=ak+1(k∈N∗ ,且 k>1)n = {a}_{k + 1}\left( {k \in {\mathrm{N}}^{ * }\text{ ,且 }k > 1}\right)n=ak+1(k∈N∗ ,且 k>1) 时,求证: bn−1≥ak⋅bn{b}_{n - 1} \geq {a}_{k} \cdot {b}_{n}bn−1≥ak⋅bn ;(ii) 求 ∑i=1Snbi\mathop{\sum }\limits_{{i = 1}}^{{S}_{n}}{b}_{i}i=1∑Snbi .