等差等比數(shù)列的綜合應(yīng)用課件

3身2019新課標(biāo)高中總復(fù)習(xí)(第1輪)XINKEBIAOGAOZHONGZONGFUXI(DI1LUN理數(shù)LISHUS湖南專用多數(shù)多圖B單元)數(shù)列與法3等盞、等丸氯列的劇多數(shù)多圖B考向預(yù)覽多數(shù)多圖B掌握等差、等比數(shù)列的基本性質(zhì):如(1)“成對(duì)”和或積相等問題;(2等差數(shù)列求和Sn1與中項(xiàng)an;能靈活運(yùn)用性質(zhì)解決有關(guān)問題如分組求和技巧、整體運(yùn)算總之,等差數(shù)列考性質(zhì),等比數(shù)列考定義。多數(shù)多圖B考點(diǎn)盤清多數(shù)多圖B1.等差數(shù)列的性質(zhì)(1)當(dāng)公差d0時(shí),等差數(shù)列的通項(xiàng)公式an=a1+(n-1)=h+a1-l是關(guān)于n的一次函數(shù),且斜率為公差的前n項(xiàng)和S0-1小的是關(guān)于n的二次函數(shù),且常數(shù)項(xiàng)為0(2)若公差①￠0,則為遞增等差數(shù)列,若公差②2d0,則為遞減等差數(shù)列,若公差③3d=0,則為常數(shù)列多數(shù)多圖B3)當(dāng)m+n=p+時(shí),則有④a+a=ata特別地,當(dāng)m+n=2p時(shí),則有an+a1(4)若{an}是等差數(shù)列,則{kan}(k是非零常數(shù)),SnS2nSnS3nS2n,…也成等差數(shù)列,而{a"n}(aAO)成等比數(shù)列;若{an}是等比數(shù)列,且an>0,則{lgan}是等差數(shù)列(5)在等差數(shù)列{an}中,當(dāng)項(xiàng)數(shù)為偶數(shù)2n時(shí)S偶S奇=③_nd;項(xiàng)數(shù)為奇數(shù)2n-1時(shí),S毒-S偶=⑥“中,S2n1=(2n1)4中(這里a中即an)奇:S偶多數(shù)多圖B(6)若等差數(shù)列{an}、{bn}的前n項(xiàng)和分別為(2n-1)aAAn、Bn,且n=fn),則Bn.(2n-1)b.B.=(2m1(7)“首正”的遞減等差數(shù)列中,前n項(xiàng)和的最大值是所有筋負(fù)項(xiàng)之和;首負(fù)”的遞增等差數(shù)列中,前n項(xiàng)和的最小是所有⑧之和(8)如果兩等差數(shù)列有公共項(xiàng),那么由它們的公共項(xiàng)順次組成的新數(shù)列也是等差數(shù)列,且新等差數(shù)列的公差是原兩等差數(shù)列公差的最小公倍數(shù)8多數(shù)多圖B2等比數(shù)列的性質(zhì)(1)若數(shù)列{an}是等比數(shù)列當(dāng)m+=+g時(shí),則有⑨aman=ann,特別地,當(dāng)m+n=2p時(shí),則有C…,·C.=2)若{an}是等比數(shù)列,則{kan}成等比數(shù)列;若{an}、{bn}成等比數(shù)列,則{anbn}、z}成等比數(shù)列;若{an}是等比數(shù)列,且公比q≠=1,則數(shù)列SnS2-Sn,S3-S2n是⑩等數(shù)列當(dāng)q=-1,且n為偶數(shù)時(shí),數(shù)列Sn,S2-Sn,S327S2n,是常數(shù)數(shù)列0,它不是等比數(shù)列多數(shù)多圖B(3)若a1>0q>1,則{an}為⑩遞增數(shù)列;若a1<0q>1,則{an}如遞減數(shù)列;若a1>00<q<1,則{an}為遞減數(shù)列;若a1<0.0<q<1,則{an為遞增數(shù)列;若φ<0,則{an}為擺動(dòng)數(shù)列;若g=1,則{an}為①常數(shù)數(shù)列4)當(dāng)q≠1時(shí),S2++