递增数列{an}满足a1=6,an+an-1=9/an-an-1+8(n>=2)则a70=

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递增数列{an}满足a1=6,an+an-1=9/an-an-1+8(n>=2)则a70=
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递增数列{an}满足a1=6,an+an-1=9/an-an-1+8(n>=2)则a70=
递增数列{an}满足a1=6,an+an-1=9/an-an-1+8(n>=2)则a70=

递增数列{an}满足a1=6,an+an-1=9/an-an-1+8(n>=2)则a70=
∵递增数列{an}满足a1=6,且an+an-1=9 /(an-an-1) +8（n≥2）,
∴an^2-an-12=8an-8an-1+9,
即 an^2-8an+16=an-12-8an-1+16+9,即 (an-4)^2=(an-1-4)^2+9,故数列{(an-4)^2}构成以9为公差的等差数列,且首项为 4．
∴(an-4)^2=4+（n-1）9=9n-5．
∴(a70-4)&2=625=25^2,
∴a70-4=25,
∴a70=29,

﹙a70﹚²－36＝8a70－8×6＋9×69
解得a70＝29 或a70＝﹣21 ∵a70≥6 ∴a70＝29

路过

∵递增数列{an}满足a1=6，且an+an-1=9 /(an-an-1) +8（n≥2），
∴an^2-an-12=8an-8an-1+9，
即 an^2-8an+16=an-12-8an-1+16+9，即 (an-4)^2=(an-1-4)^2+9，故数列{(an-4)^2}构成以9为公差的等差数列，且首项为 4．
∴(an-4)^2=4+（n-1）9=9n-5．

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∵递增数列{an}满足a1=6，且an+an-1=9 /(an-an-1) +8（n≥2），
∴an^2-an-12=8an-8an-1+9，
即 an^2-8an+16=an-12-8an-1+16+9，即 (an-4)^2=(an-1-4)^2+9，故数列{(an-4)^2}构成以9为公差的等差数列，且首项为 4．
∴(an-4)^2=4+（n-1）9=9n-5．
∴(a70-4)&2=625=25^2，
∴a70-4=25，
∴a70=29，

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