2017高三二輪講義15多角度解題數(shù)列

積微精課2017高三二輪精講15講多角度解題——數(shù)列例1、設(shè)等比數(shù)列{an}的前n項和為Sn，若S2=3，S4=15，則S6的值 q

解析：方法一：由等比數(shù)列的性質(zhì)得

=4，所以q=±2.由S2=3，解得

a-

2

a S6= = =63或S6= =

方法二：由S2，S4-S2，S6-S4成等比數(shù)列可得(S4-S2)2=S2(S6-S4)，所以 n 例2、已知Sn是等差數(shù)列{an}的前n項和，若 =4n2，則a n

n(a1an= )=

a1

n 解析：方法一：

S

=4n

2n(a1a22

2(a 2n

4n

a1a2=3所以a2=2a1，d=a2-a1=a1a3=a12d3a1=3 a n

n2方法二

=4n2

2 a5觀察發(fā)現(xiàn)可令Sn=n2+n，則an=Sn-Sn-1=n2+n-(n-1)2-(n-1)=2n，所a5

=25=5例3、記數(shù)列{an}的前n項和為Sn.若a1=1，Sn=2(a1+an)(n≥2，n∈N*)，求，{na12n≥S1+S=1122n221.，方法二：當(dāng)n=2時，S2=2(a1+a2)，從而a2=-a1=-又因為S2-2=-2，S1-2=-1，所以S2-2=2(S1-所以數(shù)列{Sn-2}是以-1為首項、2為公比的等比數(shù)列，從而Sn-2=(-1)·2n-1，所以Sn=2-2n-

例4、若等差數(shù)列a滿足a2a210，則Saa 的最大值 解法一：a2a210得a9d2a2 Saaa10a45d，所以9dS 5

積微精所以S積微精所以S2a210a的方程 100所 100，解得50S 解法二：由題可設(shè)a10cos,a10sin，公差da10a1 10sin10 Sa10a11a1910a1045d5310當(dāng)且僅當(dāng)sin1Smax

10cos50sin求證：數(shù)列an是等比數(shù)列數(shù)列b滿足b2a,b ,(n2,nN)求數(shù)列b的通 1 1在滿足（2）的條件下，求數(shù)列

cos(n1)}n項和Tn解析：（1）證明n1a1S1m1ma1，解a1n2anSnSn1man1man.即(1m)anman1m為常數(shù)，且m0，∴

m1

(n2)∴數(shù)列an是首項為1,公比（2）b12a12

1

的等比數(shù)列nb 111111(n2)n1

bn

bn1 ∴ n 2n ∴ (n1)1 2

，即bn2n1

(nN) n（3）由（2）知bn2n1，則 cos(n1) (2n1)n 所以Tn12325272 (2n1)n為偶數(shù)時Tn12523925(2n3)2n-1[322724(2n1)2n令S12523925(2n32n-4S123525927(2n72n2n3①-② 3S1242342542n1(2n3)12423(11

)(2n3)26322n33(2n3)2n126(6n13) 26(6n13)∴S9S322724(2n14S'324726(2n5)2n(2n1)③-④得3S32242442642n(2n12n124411

）(2n1)236642n43(2n1)2n228(6n7) ∴S'

28(6n7)9

SS

26(6n13)9

28(6n7)9

(6n1)2n19當(dāng)n為奇數(shù)時n1為偶數(shù)

9(6n7)2n2(18n9)

(12n2)2n

(6n1)2n1 (6n1)2n1Tn

(n為偶數(shù)9(6n1) (n為奇數(shù) 法二：Tn123252-72 (2n1)2Tn122323524725(1)n1(2n3)2n(1)n1(2n1)①-②得 3Tn1222222222 22 (2n1)223[1(