因此an?3n?7????3n?7,n?1,2,………………………………………4分
?3n?7,n?3记数列an的前n项和为Sn, 当n?1时,S1?a1?4, 当n?2时,S2?a1?a2?5,
当n?3时,Sn?S2?a3?a4???an
???5??3?3?7???3?4?7?????3n?7?
=5??n?2??2??3n?7???23211n?n?10, 22又当n?2时满足此式,
?4,n?1,?综上,Sn??3211…………………………………………8分
n?n?10,n?2?2?2(II)记数列2nan的前n项和为Tn.
则Tn?2a1?22a2?23a3???2nan,
??2Tn?22a1?23a2?24a3???2nan?1?2n?1an,
所以?Tn?2a1?d22?23???2n?2n?1an. 由(I)可知,a1??4,d?3,an?3n?7,
??41?2n?1所以?Tn??8?3???3n?7??2n?1??20??3n?10??2n?1,
1?2故Tn?20??3n?10??2n?1.………………………………………………13分 17.解:(I)记:“该射手恰好命中两次”为事件A,“该射手第一次射击甲靶命中”为事件B,“该射手第二次射击甲靶命中”为事件C,“该射手射击乙靶命中”为事件D.
由题意知,P?B??P?C????32,P?D??, 43所以P?A??PBCD?PBCD?PBCD
???????P?B?P?C?PD?P?B?PCP?D??PBP?C?P?D?
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