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已知等差数列{An},公差大于0,且A2,A5是方程x2%1...

(1)①∵等差数列{an}为递增数列,且a2,a5是方程x2-12x+27=0的两根,∴a2+a5=12,a2a5=27,∵d>0,∴a2=3,a5=9,∴d=a5?a23=2,a1=1,∴an=2n-1(n∈N*)②∵Tn=1-12bn,∴令n=1,得b1=23,当n≥2时,Tn=1-12bn,Tn-1=1-12bn-1,两式相减得,bn=12bn-1-1...

∵a2,a8是方程x2-12x+m=0的两根,∴a2+a8=12,a2a8=m.∵{an}的前15项的和为5m,∴15(a1+a15)2=5m,∴3a8=m,联立a2+a8=12a2a8=m3a8=m,解得a2=12a8=0或a2=3a8=9.∵对任意正整数n,都有an>an+1,∴取a2=12a8=0.∴0=12+(8-2)d,解得d=-2....

(1)an=(n+2)/2 (2)Sn=2-(n+4)/2^(n+1)

因为关于x的不等式dx2+2a1x≥0的解集为[0,9],所以d<0,且81d+18a1=0,解得a1=?92d,故an=a1+(n-1)d=(n-112)d,令(n-112)d≤0,(注意d<0),解得n≥112,即等差数列{an}的前5项为正,从第6项开始为负,故数列{an}的前5项和S5取最大,故答...

x∧2-5x+6=0即为(x-2)(x-3)=0,所以x=2或3,因为是等差数列,a2>a4,所以a2=3,a4=2,a2-2d=a4,所以d=0.5,a1=3.5,所以等差数列是:an=3.5-(n-1)*0.5

(1)设等差数列{an}的首项为a1,公差为d∵a2=5,a4+a6=22,∴a1+d=5,2a1+8d=22,解得a1=3,d=2,∴an=2n+1,Sn=n2+2n.(2)∵f(x)=1x2?1,bn=f(an),∴bn=1an2?1,∵an=2n+1∴an2?1=4n(n+1),∴bn=14n(n+1)=14(1n?1n+1),Sn=b1+b2+…+bn=14(1?12+...

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(1)设数列{an}公差为d,则a1+a2+a3=3a1+3d=6,又a1=1,∴d=1.所以an=n.(2)解:令Sn=b1+b2+…+bn,则由bn=anxn=nxn,得Sn=x+2x2+…(n?1)xn?1+nxn,①xSn=x2+2x3+…+(n?1)xn+nxn+1,②当x≠1时,①式减去②式,得(1?x)Sn=(x+x2+…xn)?nxn+1=x(1?...

(Ⅰ)证明:由题意f(an)=4+(n-1)×2=2n+2,即logkan=2n+2,(1分)∴an=k2n+2∴an+1an=k2(n+1)+2k2n+2=k2.(2分)∵常数k>0且k≠1,∴k2为非零常数,∴数列{an}是以k4为首项,k2为公比的等比数列.(3分)(II)解:由(1)知,bn=anf(an)=k2...

(1)∵点(a8,bn)在函数f(x)=2x的图象上,∴bn=2an,又等差数列{an}的公差为d,∴bn+1bn=2an+12an=2an+1?an=2d,∵点(a8,4b7)在函数f(x)的图象上,∴4b7=2a8=b8,∴b8b7=4=2d,解得d=2.又a1=-2,∴Sn=na1+n(n?1)2d=-2n+n(n?1)2×2=n2-3n....

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