【答案】
分析:(I)由已知利用遞推公式
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213226335948300/SYS201310232132263359483019_DA/0.png)
可得a
n,代入分別可求數(shù)列b
n的首項b
1,公比q,從而可求b
n(II)由(I)可得c
n=(2n-1)•4
n-1,利用乘“公比”錯位相減求和.
解答:解:(1):當(dāng)n=1時,a
1=S
1=2;當(dāng)n≥2時,a
n=S
n-S
n-1=2n
2-2(n-1)
2=4n-2,
故{a
n}的通項公式為a
n=4n-2,即{a
n}是a
1=2,公差d=4的等差數(shù)列.
設(shè){b
n}的通項公式為q,則b
1qd=b
1,d=4,∴q=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213226335948300/SYS201310232132263359483019_DA/1.png)
.
故b
n=b
1q
n-1=2×
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213226335948300/SYS201310232132263359483019_DA/2.png)
,即{b
n}的通項公式為b
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213226335948300/SYS201310232132263359483019_DA/3.png)
.
(II)∵c
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213226335948300/SYS201310232132263359483019_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213226335948300/SYS201310232132263359483019_DA/5.png)
=(2n-1)4
n-1,
T
n=c
1+c
2+…+c
nT
n=1+3×4
1+5×4
2+…+(2n-1)4
n-14T
n=1×4+3×4
2+5×4
3+…+(2n-3)4
n-1+(2n-1)4
n兩式相減得,3T
n=-1-2(4
1+4
2+4
3+…+4
n-1)+(2n-1)4
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213226335948300/SYS201310232132263359483019_DA/6.png)
[(6n-5)4
n+5]
∴T
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213226335948300/SYS201310232132263359483019_DA/7.png)
[(6n-5)4
n+5]
點評:(I)當(dāng)已知條件中含有s
n時,一般會用結(jié)論
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213226335948300/SYS201310232132263359483019_DA/8.png)
來求通項,一般有兩種類型:①所給的s
n=f(n),則利用此結(jié)論可直接求得n>1時數(shù)列{a
n}的通項,但要注意檢驗n=1是否適合②所給的s
n是含有a
n的關(guān)系式時,則利用此結(jié)論得到的是一個關(guān)于a
n的遞推關(guān)系,再用求通項的方法進行求解.
(II)求和的方法的選擇主要是通項,本題所要求和的數(shù)列適合乘“公比”錯位相減的方法,此法是求和中的重點,也是難點.