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1、數(shù)列求和的常見類型和方法一、公式法：1等差等比數(shù)列求和直接用公式.21+2+3+…+n=n(n+1)2,12+22+32+…+n2=16nn+12n+1,13+23+33+…+n3=n(n+1)22,1+3+5+…+2n-1=n2.3Cn0+Cn1+Cn2+…+Cnn=2n,Cn0+Cn2+Cn4+…=Cn1+Cn3+Cn5+…=2n-1Crr+Cr+1r+…+Cnr=Cn+1r+1Cnk+Cnk+1=Cn+1k+1,Arr+Ar+1r+…+Anr=Cn+1r+1AkkAnk=Cnk?AkkCn0+12Cn1+13Cn2+…+1n+1Cnn=2n+1-1n+11k+1C
2、nk=1n+1Cn+1k+10Cn0+1Cn1+2Cn2+…+nCnn=n2n-1(4)已知Fan,Sn=0求an.有以下兩種常見途徑：途徑一：Fan,Sn=0?Sn=fanSn-1=fan-1?an=fan-fan-1?an?Sn(n≥2)途徑二：Fan,Sn=0?an=fSn?Sn-Sn-1=fSn?Sn(n≥2)二、裂項法：11等差×等差求和用裂項法.設數(shù)列an為等差數(shù)列an≠0,d≠0,則：1a1a2+1a2a3+…+1anan+1=1d1a1-1an+11anan+1=1d(1an-1an+1)211?2?3+12?3?4+…+1
3、nn+1n+2=1212-1n+1n+21nn+1(n+2)=121nn+1-1n+1(n+2)注：可推廣為:設數(shù)列an為等差數(shù)列an≠0,d≠0,由于1anan+1an+2=12d(1anan+1-1an+1an+2)則：1a1a2a3+1a2a3a4+…+1anan+1an+2=12d1a1a2-1an+1an+2而且還可以進一步推廣.311+2+12+3+…+1n+n+1=n+1-11n+n+1=-n+n+1注：可推廣為:設數(shù)列an為等差數(shù)列an>0,d>0,由于1an+an+1=1d-an+an+1,41則：1a1+a2+1a2+a3+…+1an+an+1=1d
4、-a1+an+1.41A22+1A32+…+1An2=1-1n1An2=1n-1-1n,1C22+1C32+…+1Cn2=2-2n1Cn2=2(1n-1-1n)12!+23!+34!+…+nn+1!=1-1n+1!nn+1!=n+1-1n+1!=1n!-1n+1!1?1!+2?2!+3?3!+…+n?n!=n+1!-1n?n!=n+1-1n!=n+1!-n!52121-122-1+2222-1(23-1)+…+2n2n-1(2n+1-1)=1-12n+1-12n2n-1(2n+1-1)=12n-1-12n+1-1總結(jié)：裂項法求和的本質(zhì)就是:將數(shù)列的每一項裂成另一數(shù)列相鄰
5、兩項之差,造成相消項,從而達到化簡求和的目的,即Sn=i=1nai=i=1nbi+1-bi=bn+1-b1.三、形如“a1b1+a2b2+a3b3+…+anbn”求和：1等差×等差若an等差,bn等差,則用分組求和法.設an=k1n+b1,bn=k2n+b2,∵anbn=k1n+b1k2n+b2=k1k2n2+k1b2+k2b1n+b1b2∴Sn=k1k212+22+…+n2+k1b2+k2b11+2+…+n+b1b2n=…注：可推廣為:“等差×等差×等差”甚至更多等差之積求和用分組求和法.2等差×等比若an等差,bn等比,則用錯位相減法.設an=kn+b,bn=b1q
6、n-1(q≠1),∵Sn=a1b1+a2b2+…+anbn①∴qSn=a1b1q+…+an-1bn-1q+anbnq=a1b2+…+an-1bn+anbn+1②①-②得:1-qSn=a1b1+db2+b3+…+bn-anbn+1=a1b1+db21-qn-11-q-anbn+1?Sn=a1b1+db21-qn-11-q-anbn+11-q=…41注：可推廣為:“等差×等差×等比”甚至更多等差×等比求和用錯位相減法(用兩次(甚至多次)錯位相減法).四、形如“a1Cn0+a2Cn1+a3Cn2+…+an+1Cnn”求和：1等差×組合數(shù)若an等差,則用倒序相加法.設公差為d∵
7、Sn=a1Cn0+a2Cn1+…+anCnn-1+an+1Cnn①∴Sn=an+1Cnn+anCnn-1+…+a2Cn1+a1Cn0=an+1Cn0+anCn1+…+a2Cnn-1+a1Cnn②∴①+②:2Sn=a1+an+1Cn0+Cn1+…+Cnn-1+Cnn=2a1+nd2n?Sn=2a1+nd2n-1.2等比×組合數(shù)若an等比,則逆用二項式定理.設公比為q∴Sn=a1Cn0+a1qCn1+a2qCn2+…+a1qCnn=a1Cn0+Cn1q+Cn2q2+…+Cnnqn=a1(1+q)n五、數(shù)歸法：六、構造法：1Cn02+Cn12+
