2025屆高考數(shù)學(xué)二輪總復(fù)習(xí)專題3數(shù)列專項(xiàng)突破3數(shù)列解答題課件

專項(xiàng)突破三數(shù)列解答題考點(diǎn)一等差、等比數(shù)列的判定與證明例1(2021全國甲,理18)已知數(shù)列{an}的各項(xiàng)均為正數(shù),記Sn為{an}的前n項(xiàng)和,從下面①②③中選取兩個(gè)作為條件,證明另外一個(gè)成立.①數(shù)列{an}是等差數(shù)列;②數(shù)列

是等差數(shù)列;③a2=3a1.∴a2=a1+d1=3a1.若選①③?②設(shè)等差數(shù)列{an}的公差為d.∵a2=3a1,∴a1+d=3a1,∴d=2a1,=n2a1-(n-1)2a1=(2n-1)a1,當(dāng)n=1時(shí),an=(2n-1)a1也成立,∴an=(2n-1)a1,n∈N*.又an+1-an=(2n+1)a1-(2n-1)a1=2a1(常數(shù)),∴數(shù)列{an}是等差數(shù)列.[對點(diǎn)訓(xùn)練1](2024山東高中名校統(tǒng)一調(diào)研)已知數(shù)列{an},{bn}是公比不相等的兩個(gè)等比數(shù)列,令cn=an+bn.(1)證明:數(shù)列{cn}不是等比數(shù)列;(2)若an=2n,bn=3n,是否存在常數(shù)k,使得數(shù)列{cn+1+kcn}為等比數(shù)列?若存在,求出k的值;若不存在,說明理由.整理得12(2+k)(3+k)=13(2+k)(3+k),解得k=-2或k=-3.經(jīng)檢驗(yàn),當(dāng)k=-2時(shí),cn+1+kcn=2n+1+3n+1+(-2)·(2n+3n)=3n,此時(shí)數(shù)列{cn+1-2cn}為等比數(shù)列;當(dāng)k=-3時(shí),cn+1+kcn=2n+1+3n+1+(-3)·(2n+3n)=-2n,數(shù)列{cn+1-3cn}為等比數(shù)列,所以存在常數(shù)k=-2或k=-3,使得數(shù)列{cn+1+kcn}為等比數(shù)列.考點(diǎn)二證明數(shù)列不等式[對點(diǎn)訓(xùn)練2](2024廣東珠海模擬)已知數(shù)列{an}的前n項(xiàng)和為Sn,且2Sn=(n+2)(an+1).(1)求數(shù)列{an}的通項(xiàng)公式;(1)解

由2Sn=(n+2)(an+1),則2Sn+1=(n+3)(an+1+1),則2Sn+1-2Sn=(n+3)(an+1+1)-(n+2)(an+1)=2an+1,考點(diǎn)三求數(shù)列不等式中參數(shù)的范圍(最值)問題例3(2024福建泉州高三檢測)已知正項(xiàng)數(shù)列{an},{bn}滿足:對任意正整數(shù)n,都有an,bn,an+1成等差數(shù)列,bn,an+1,bn+1成等比數(shù)列,且a1=10,a2=15.(方法一

參變不分離)設(shè)f(n)=(λ-1)n2+(3λ-6)n-8,n∈N*,則f(n)<0對任意正整數(shù)n恒成立,當(dāng)λ-1>0,即λ>1時(shí),不滿足條件;當(dāng)λ-1=0,即λ=1時(shí),f(n)=-3n-8,滿足條件;當(dāng)λ-1<0,即λ<1時(shí),令f(x)=(λ-1)x2+(3λ-6)x-8,綜上,當(dāng)實(shí)數(shù)λ≤1時(shí),不等式恒成立.[對點(diǎn)訓(xùn)練3]數(shù)列{an}中,a1=8,a4=2,且滿足an+2=2an+1-an,n∈N*.(1)求a2,a3,并求數(shù)列{an}的通項(xiàng)公式;解

(1)由an+2=2an+1-an,n∈N*知數(shù)列{an}為等差數(shù)列,設(shè)其公差為d.令n=1,a3=2a2-a1=2a2-8,令n=2,a4=2a3-