设 {an}\left\{ {a}_{n}\right\}{an} 是等差数列, {bn}\left\{ {b}_{n}\right\}{bn} 是等比数列. 已知 a1=4,b1=6,b2=2a2−2,b3=2a3+4{a}_{1} = 4,{b}_{1} = 6,{b}_{2} = 2{a}_{2} - 2,{b}_{3} = 2{a}_{3} + 4a1=4,b1=6,b2=2a2−2,b3=2a3+4 .(1) 求 {an}\left\{ {a}_{n}\right\}{an} 和 {bn}\left\{ {b}_{n}\right\}{bn} 的通项公式;(2) 设数列 {cn}\left\{ {c}_{n}\right\}{cn} 满足 c1=1,cn={1, 2k<n<2k+1,bk,n=2k,{c}_{1} = 1,{c}_{n} = \left\{ \begin{array}{l} 1,\;{2}^{k} < n < {2}^{k + 1}, \\ {b}_{k},n = {2}^{k}, \end{array}\right.c1=1,cn={1,2k<n<2k+1,bk,n=2k, 其中 k∈N∗k \in {\mathbf{N}}^{ * }k∈N∗ .(i) 求数列 {a2n(c2n−1)}\left\{ {{a}_{{2}^{n}}\left( {{c}_{{2}^{n}} - 1}\right) }\right\}{a2n(c2n−1)} 的通项公式;(ii) 求 ∑i=1na2ic2i (n∈N∗)\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{{2}^{i}}{c}_{{2}^{i}}\;\left( {n \in {\mathbf{N}}^{ * }}\right)i=1∑na2ic2i(n∈N∗) .