6、}⑵求數(shù)列a,b的通項a和b;⑶設cab,求數(shù)列c的前n項和To高三二輪復習4、數(shù)列{aj的通項公式2013-3-27,若前n項的和為10,則項數(shù)為n+n+1A.11B.99C?120D.1211Vn+Vn~l{ajfiSSj=^(2*-l)”=2411>數(shù)列’的前〃項和為J'八,"為正整數(shù).若°,則解析van=yln+1—Vn,??S=*71+1—1=10,..n=120.a4SS15a4317a1=8=24二a1_3a+112.已知{an}是各項為正數(shù)的等比數(shù)列,一個等比中項.則數(shù)列是各項為正數(shù)的盂比數(shù)列,且+rn+=2^2848385+/.a2
7、2aai2{an}的通項公気8l9T32.32842lol,8385TOOa_2'aa24=(□2=a-=_2J2a)=*100即:'4aJ04_21642時,2(時,100+解:舍去),a28213、已知等比數(shù)列"{aj的首項為1?若4a?i,2a2,83成等差數(shù)列,則數(shù)列的前5項和為解析設數(shù)列{a』的公比為q,v4ai,2a2?11???數(shù)列是首項=1an1,ai14、對予數(shù)列{In},定義數(shù)列口,則數(shù)列列”的通項為2n解析van+i—an=2,,4是a2和34的(1)7a}+=84a_2qan10a8或22a4na2q3=an2a2qa3成等差
8、數(shù)列,???4q=4+q1公比為的等比尊列,???1+211+816舍去)4解得q=2,31=16{an+i-an}為數(shù)列{aj的“差數(shù)列”,若ai=2,{aj的嗟數(shù){an}的前n項和S=n—1n—2.an=(an—an-i)+(an-i—an-2)H—+(a2—ad+a〔=2+2H—+2=2—2—+"2+2=2:2-2n+1??S=[一2_2一Z其中m為常數(shù),且15、設數(shù)列{aj的前n項和Sn,且(3m)Sn2mam3(nN*)nm3,m0.(I)求值d是等比數(shù)列;3(U)若數(shù)列{an}的公比q=f(m),數(shù)列{b』滿足bl=a=f(bn-i)(
9、nEN,n~2),12求證{JL}為等差數(shù)列,并求bnb+(3得)_12mamSn【答案】(I)由(3)S2