(Ⅰ)求数列?an?和?bn?的通项公式； (Ⅱ)设cn?anbn，求数列?cn?前n项和Tn．
22解：(Ⅰ)?当n?1时，当n?2时， an?Sn?Sn?1?2n?2?n?1??4n?2，a1?S1?2；
故?an?的通项公式为an?4n?2,即?an?是首项为2，公差为4的等差数列. (Ⅱ) ?cn?anbn?4n?224n?1??2n?1?4n?1，
?Tn?c1?c2???cn ?1?3?4?5?4????2n?1?41212n?1
??2n?1?4n4Tn? 1?4?3?4????2n?3?4n?1两式相减得
3Tn??1?2?4?4???412n?1???2n?1?4?n1??6n?5?4n?5??3?
?Tn?1??6n?5?4n?5? ?9?