求数列通项公式的十种方法.docx

WOIRD格式:..专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式专业资料整理WOIRD格式解法二:由题设nSnS○1nn1(3)(n1)Sn(n2)Sn○21①的两边分别减去②的两边,整理得na1(n2)a,,323a5a,,,43(n1)a(n1)a,,得(n1)!(n1)!aa,n26n(n1)n(n1)由(Ⅰ)并化简得aa2,,,所以nnn22b1b(n2)(n1),nn即bbnn1221,(n1)(n2)*(n1),则xnxn11,,,即(n1)2b(n1),:由题设有nS1(n3)S,nn*nN,所以S24S1,2S35S2,,,(n1)Sn(n2)Sn1,,得12(n1)Sn45(n2)S,化简得1n(n1)(n2)n(n1)(n2)Sa,(Ⅰ),上式对n1,(n1)aSS,,可得2b(n1),(Ⅲ)证明:n(n1)aaa222T(1)b(1)b(1)b23(1)(n1).212nn12n当n4k,*kN时,2222(42)2222Tn2345k(4k1)(4k)(4k1).注意到2222(4k2)(4k1)(4k)(4k1)32k4,专业资料整理WOIRD格式故Tnk(k1)32(12k)43242224k(4k4)4(4k),*kN时,24(41)24(1)24(1)(2)24Tnkkknnnn(4)421当n4k2,*kN时,24(41)(4)44(2)(3)4224222Tn(4k),*kN时,(41)(41)4(3)(4)(2)44*4, n4k3,kN所以Tn2nn nkk24,42,2n1,n4k1,k *NN*.2nn4,n4k,kN *42n0,n5,9,13,从而n3时,有2412,n6,10,14,|T|nn2n2n212,n3,7,11,2nn1412,n4,8,12,2nn总之,当n3时有|T|n2n2,即2|Tn|