2020屆高考數(shù)學(xué) 數(shù)列部分新創(chuàng)題
(2)數(shù)列部分新創(chuàng)題4道1若等比數(shù)列an對(duì)一切正整數(shù)n都有Sn=2an-1,其中 Sn是an的前n項(xiàng)和,則公比q的值為 ( )A. B.- C.2 D.-22. 等差數(shù)列an的通項(xiàng)公式是an=1-2n,其前n項(xiàng)和為Sn,則數(shù)列的前11項(xiàng)和為 ( )A.-45 B.-50 C.-55 D.-663. 等差數(shù)列an中有兩項(xiàng)am和ak滿足am=,ak=,則該數(shù)列前mk項(xiàng)之和是 .4. 設(shè)f(x)= (a>0)為奇函數(shù),且 |f(x)|min=2,數(shù)列an與bn滿足如下關(guān)系:a1=2,an+1=.(1)求f(x)的解析表達(dá)式;(2)證明:當(dāng)nN+時(shí),有bn()n.參考答案:1C 當(dāng)n=1時(shí),S1=2a1-1,得a1=1;當(dāng)n=2時(shí),1+a2=2a2-1,得公比q=a2=a1q=2.2. DSn=,=-n,前11項(xiàng)的和為-66.3. 設(shè)數(shù)列an的首項(xiàng)為a1,公差為d,則有解得,所以Smk=(a1+am)=.4.解:(1)由f(x)是奇函數(shù),得b=c=0,由|f(x)|min=2,得a=2,故f(x)=.(2)an+1=bn=b,而b1=,bn=()2n-1.當(dāng)n=1時(shí),b1=,命題成立;當(dāng)n2時(shí),2n-1=(1+1)n-1=1+C+C+C1+C=n,()2n-1()n,即bn()n.