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已知{An}是各项都为正数的等比数列,Sn是其前n项和...

∵数列{an}为各项都是正数的等比数列,设公比为q,则q>0,由已知数据可知q≠1,∴S10=a11?q(1-q10)=10,①S30=a11?q(1-q30)=70,②①②两式相除可得q20+q10+1=7,解得q10=2或q10=-3(舍去)把q10=2代入①可得a11?q=-10,∴S40=a11?q(1-q40)=-10×(...

由题意可设{an}的公比为q,q>0∵5S2=S4,∴5(a1+a2)=a1+a2+a3+a4,∴4(a1+a2)=a3+a4,即4(a1+a2)=(a1+a2)q2,∴q2,=4,解得q=2,∴a5=a1q4=1×24=16故选:C

设等比数列{an}的公比为q,(q>0)由题意可得2×12a3=3a1+2a2,即a1q2=3a1+2a1q,即q2-2q-3=0解之可得q=3,或q=-1(舍去)故S11?S9S7?S5=a10+a11a6+a7=a6q4+a7q4a6+a7=q4=81故答案为:81

(I)设{an}的公差为d,{bn}的公比为q,则依题意有q>0,∵a1=b1=1,a3+b5=21,a5+b3=13,∴1+2d+q4=211+4d+q2=13,解得d=2,q=2.∴an=1+(n-1)d=2n-1,bn=2n?1,(Ⅱ)由(I)得,an?bn=(2n-1)?2n-1,Sn=1?20+3?21+…+(2n-1)?2n-12Sn=1?2+3...

前三项设为a1,a1q,a1q^2 前三项和为21 a1+a1q+a1q^2=21 a1=3带入 得 1+q+q^2=7 q=2或-3(舍) an=a1*q^n=3*2^n(n为正整数)

(1)解:由a2n-1,a2n,a2n+1成等差数列,可知a1,a2,a3成等差数列,a3,a4,a5成等差数列,由a2n,a2n+1,a2n+2成等比数列,可知a2,a3,a4成等比数列,则2=a1+a3a32=a2a42a4=a3+a5,又a2=1,a5=3,∴2=a1+a3a32=a42a4=a3+3,则2a32=a3...

设a2=a1*q,a3=a1*q^2 a1(1+q^2)=10, a1q(1+q)=6 (1+q^2)/(q+q^2)=10/6=5/3 跳过几步,可以解得 a1=8,q=1/2

S1=3a1-1 ===>a1=1/2 an=Sn-S(n-1) =3an-1-3a(n-1)+1 化简an/a(n-1)=3/2 所{an}是以a1=1/2为首相q=3/2为公比的等比数列 an=(1/2)*(3/2)^(n-1)

(1)设an的公差为d,bn的公比为q,则依题意有q>0且1+d+q2=71+2d+q=7解得d=2,q=2.(2分)所以an=1+(n-1)d=2n-1,bn=qn-1=2n-1.(4分)(2)因为cn=an-2010=2n-2011≥0?n≥1005.5,所以,当1≤n≤1005时,cn<0,当n≥1006时,cn>0.(6分)...

a4*a10=(a7)^2=(2a6)^2=16 所以 a6=2

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