適用于新高考新教材廣西專(zhuān)版2024屆高考數(shù)學(xué)二輪總復(fù)習(xí)專(zhuān)題3數(shù)列素養(yǎng)提升微專(zhuān)題四數(shù)列解答題中的奇偶項(xiàng)問(wèn)題課件

素養(yǎng)提升微專(zhuān)題(四)數(shù)列解答題中的奇、偶項(xiàng)問(wèn)題第一編規(guī)律方法數(shù)列中的奇、偶項(xiàng)問(wèn)題是對(duì)一個(gè)數(shù)列分成兩個(gè)新數(shù)列進(jìn)行單獨(dú)研究,利用新數(shù)列的特征(等差、等比數(shù)列或其他特征)求解原數(shù)列.數(shù)列中奇、偶項(xiàng)問(wèn)題的常見(jiàn)題型有:(1)數(shù)列中連續(xù)兩項(xiàng)的和或積的問(wèn)題(an+an+1=f(n)或an·an+1=f(n));(2)通項(xiàng)中含有(-1)n的類(lèi)型;(3){a2n},{a2n-1}的類(lèi)型;(4)已知條件中明確的奇、偶項(xiàng)問(wèn)題.考查角度角度一

通項(xiàng)中含有(-1)n的數(shù)列求和[例1]已知正項(xiàng)數(shù)列{an},其前n項(xiàng)和為Sn,an=1-2Sn.(1)求數(shù)列{an}的通項(xiàng)公式;角度二

奇、偶項(xiàng)通項(xiàng)不同的數(shù)列求和[例2](2023·廣東河源高三期末)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a5=5,S8=36,an=log3bn.(1)求數(shù)列{an}和{bn}的通項(xiàng)公式;解

(1)設(shè)等差數(shù)列{an}的公差為d.∵a5=5,S8=36, ∴an=1+(n-1)×1=n.又an=log3bn,∴n=log3bn,∴bn=3n.對(duì)點(diǎn)演練1.

已知數(shù)列{an}的前n項(xiàng)和為Sn,且2Sn=3an-3.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=2log3an+(-1)n·n,求數(shù)列{bn}的前n項(xiàng)和Tn.解

(1)當(dāng)n=1時(shí),2S1=2a1=3a1-3,所以a1=3.因?yàn)?Sn=3an-3,所以當(dāng)n≥2時(shí),2Sn-1=3an-1-3,所以數(shù)列{an}是首項(xiàng)為3,公比為3的等比數(shù)列,所以an=3n.2.設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,且a2=8,S4=40.數(shù)列{bn}的前n項(xiàng)和為T(mén)n,且Tn-2bn+3=0.(1)求數(shù)列{an},{bn}的通項(xiàng)公式;因?yàn)門(mén)n-2bn+3=0,所以當(dāng)n=1時(shí),b1-2b1+3=0,即b1=3;當(dāng)n≥2時(shí),Tn-1-2bn-1+3=0,所以Tn-Tn-1-2bn+2bn-1=-bn+2bn-1=0,即bn=2bn-1(n≥2).所以數(shù)列{bn}是首項(xiàng)為3,公比為2的等比數(shù)列,所以bn=3×