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已知等差数列{An},公差大于0,且A2,A5是方程x2%1...

(Ⅰ)设an的首项为a1,∵a2,a5是方程x2-12x+27=0的两根,∴a2+a5=12a2?a5=27?a1=1d=2∴an=2n-1n=1时,b1=T1=1?12b1∴b1=23n≥2时,Tn=1?12bn,Tn?1=1?12bn?1,两式相减得bn=13bn?1数列是等比数列,∴bn=23?(13)n?1(Ⅱ)cn=(2n?1)?23?(13)...

(1)∵等差数列{an}的公差d>0,a2、a5且是方程x2-12x+27=0的两根,∴a2=3,a5=9.∴d=9?35?2=2,∴an=a2+(n-2)d=3+2(n-2)=2n-1;又数列{bn}中,Tn=1-12bn,①∴Tn+1=1-12bn+1,②②-①得:bn+1bn=13,又T1=1-12b1=b1,∴b1=23,∴数列{bn}是以23为首...

解答:解 (1)∵a3,a5是方程x2-14x+45=0的两根,且数列{an}的公差d>0,解方程x2-14x+45=0,得x1=5,x2=9,∴a3=5,a5=9,┅(2分)∴a1+2d=5a1+4d=9,解得a1=1d=2┅(4分)∴an=a1+(n-1)d=2n-1┅(6分)(2)∵an=2n-1,∴bn=2an+n=22n?1+n┅(...

(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...

(1)由方程x2-14x+45=0,解得x=5或9.∵a3和a5是方程x2-14x+45=0的两根,且公差d>0.∴a3=5,a5=9,∴a1+2d=5a1+4d=9,解得a1=1d=2.∴an=1+2(n-1)=2n-1.由Sn=1?bn2(n∈N*),当n=1时,b1=S1=1?b12,解得b1=13.当n≥2时,bn=Sn-Sn-1=1?bn2?...

(Ⅰ)∵a3,a5是方程x2-14x+45=0的两根,且数列{an}的公差d>0,解得a3=5,a5=9,则公差d=a5?a35?3=2.∴an=a5+(n-5)d=2n-1.当n=1时,有b1=S1=1-12b1,∴b1=23,当n≥2时,有bn=Sn-Sn-1=12(bn?1?bn),∴3bn=bn-1,∵b1=23≠0,∴bnbn?1=13(n≥2)...

(1)∵等差数列{an}的公差大于0,且a2,a4是方程x2-14x+45=0的两根,∴a2<a4,解方程x2-14x+45=0,得:a2=5,a4=9,∴a1+d=5a1+3d=9,解得a1=3,d=2,∴an=3+(n-1)×2=2n+1.∵数列{bn}的前n项的和为Sn,且bn+Sn=1,∴Sn=1-bn,n=1时,b1=1-b1,...

(1)∵a3,a5分别是方程x2-14x+45=0的两个实根,d>0∴a3<a5∴a3=5a5=9 an=5+2(n?3)=2n?1(n∈N*)(2)由(1)可得,bn=2n2n+1=n(12)nTn=12+2?(12)2+3?(12)3+…+n?(12)n∴12Tn=(12)2+2?(12)3+…+(n?1)?(12)n+n?(12)n+1两式相减可得,12Tn=12+(1...

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

(1)解方程x2-12x+27=0,可得x=3或9,∵a2、a5是方程x2-12x+27=0的两根,数列{an}是递增的等差数列,∴a2=3,a5=9,设公差为d,则a1+d=3a1+4d=9,解得a1=1,d=2,∴an=1+2(n-1)=2n-1.对于数列{bn},Sn=1-12bn(n∈N+).当n=1时,b1=1?12b1,...

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