数列求和的常见类型和方法.docx

数列求和的常见类型和方法一、公式法：1等差等比数列求和直接用公式.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=CnkAkkCn0+12Cn1+13Cn2+1n+1Cnn=2n+1-1n+11k+1Cnk=1n+1Cn+1k+10Cn0+1Cn1+2Cn2+nCnn=n2n-1(4)已知Fan,Sn=0求an.有以下两种常见途径：途径一：Fan,Sn=0Sn=fanSn-1=fan-1an=fan-fan-1anSn(n2)途径二：Fan,Sn=0an=fSnSn-Sn-1=fSnSn(n2)二、裂项法：11等差等差求和用裂项法.设数列an为等差数列an0,d0,则：1a1a2+1a2a3+1anan+1=1d1a1-1an+11anan+1=1d(1an-1an+1)21123+1234+1nn+1n+2=1212-1n+1n+21nn+1(n+2)=121nn+1-1n+1(n+2)注：可推广为:设数列an为等差数列an0,d0,由于1anan+1an+2=12d(1anan+1-1an+1an+2)则：1a1a2a3+1a2a3a4+1anan+1an+2=12d1a1a2-1an+1an+2而且还可以进一步推广.311+2+12+3+1n+n+1=n+1-11n+n+1=-n+n+1注：可推广为:设数列an为等差数列an0,d0,由于1an+an+1=1d-an+an+1,则：1a1+a2+1a2+a3+1an+an+1=1d-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!11!+22!+33!+nn!=n+1!-1nn!=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总结：裂项法求和的本质就是:将数列的每一项裂成另一数列相邻两项之差,造成相消项,从而达到化简求和的目的,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,则用分组求和法.设an=k1n+b1,bn=k2n+b2,anbn=k1n+b1k2n+b2=k1k2n2+k1b2+k2b1n+b1b2Sn=k1k212+22+n2+k1b2+k2b11+2+n+b1b2n=注：可推广为:“等差等差等差”甚至更多等差之积求和用分组求和法.2等差等比若an等差,bn等比,则用错位相减法.设an=kn+b,bn=b1qn-1(q1),Sn=a1b1+a2b2+anbnqSn= a1b1q+an-1bn-1q+anbnq = a1b2+an-1bn+anbn+1 -得:1-qSn=a1b1+db2+b3+bn-anbn+1 =a1b1+db21-qn-11-q-anbn+1Sn=a1b1+db21-qn-11-q-anbn+11-q=注：可推广为:“等差等差等比”甚至更多等差等比求和用错位相减法(用两次(甚至多次)错位相减法).四、形如“a1Cn0+a2Cn1+a3Cn2+an+1Cnn”求和：1等差组合数若an等差,则用倒序相加法.设公差为dSn=a1Cn0+a2Cn1+anCnn-1+an+1CnnSn=an+1Cnn+anCnn-1+a2Cn1+a1Cn0 =an+1Cn0+anCn1+a2Cnn-1+a1Cnn +:2Sn=a1+an+1Cn0+Cn1+Cnn-1+Cnn=2a1+nd2nSn=2a1+nd2n-1.2等比组合数若an等比,则逆用二项式定理.设公比为qSn=a1Cn0+a1qCn1+a2qCn2+a1qCnn =a1Cn0+Cn1q+Cn2q2+Cnnqn=a1(1+q)n五、数归法：六、构造法：1Cn02+Cn12+Cnn2=C2nn,考查恒等式(1+x)n(1+x)n=(1+x)2n两边xn的系数.2Cn0Cmm+Cn1Cmm-1+CnmCm0=Cm+nmnm,考查恒等式(1+x)m(1+x)n=(1+x)m+n两边xm的系数.七、特殊方法：在二项式展开式中,有关系数数列的求和常有以下一些特殊方法:赋值法,导数法等. 设(ax+b)n=a0+a1x+a2x2+a3x3+anxn在中,令x=1可得:a0+a1+a2+a3+an=a+bn在中,令x=-1可得:a0-a1+a2-a3+(-1)nan=-a+bn联立,可解得:a0+a2+a4+=a+bn+-a+bn2 a1+a3+a5+=a+bn-a+bn2由,还可得:a0+a1+a2+a3+an=a+bn对两边求导得:anax+bn-1=a