解得a1=2,由4Sn=an2+2an,n用n-1代,两式相减得an2-an-12=2(an+an-1),得an=2n.对n=1也成立.则数列{an}的通项公式为an=2n;(Ⅱ)bn=2n•3n,错位相减法可以得Sn=2•3+4•32+…+2n•3n,3Sn=2•32+4•33+…+2n•3n+1,两式相减可得,-2Sn=2(3+32+…+3n)-2n...
令n=1 S1=a1由an^2+2an=4Sn.(1)解得a1=24Sn-1=an-1^2+2an-1.(2)(1)-(2)得4an=an^2-an-1^2+2an-2an-1化简得（an+an-1)（an-an-1-2）=0.an+an-1≠0 则an-an-1-2=0an=an-1+2{an}是首项a1=2公差d=2 的等差数列an=2n...
【解析】根据题意得an2+2an=4Sn+3①当n=1时,a12+2a1=4S1+3即a12-2a1-3=0∴(a1+1(a1-3)=0解得a1=3或a1=-1(舍去)②当n≥2时,则有an-12+2an-1=4Sn-1+3又an2+2an=4Sn+3∴(an2+2an)-(an-2+2an-1)=(4Sn+3)-(4Sn-1+3). an-an-1=+2an-1 ∵an0.∴an-an-1=2{an...
解答 解:(1)由题意得{a2n+2an=4Sna2n+1+2an+1=4Sn+1{an2+2an=4Snan+12+2an+1=4Sn+1,两式作差得(an+1+an)(an+1-an-2)=0,又数列{an}的各项都为正数,∴an+1-an-2=0,即an+1-an=2,当n=1时,有a21+2a1a12+2a1=4a1,解得a1=2,∴数列{an}为等差数列,首项为2,公差为2.∴S...
an-1^2+2an-1=4Sn-1 做差：(an-an-1)(an+an-1)+2(an-an-1)=4(Sn-Sn-1)=4an (an-an-1)(an+an-1)-2(an+an-1)=0 (an-an-1-2)(an+an-1)=0 an>0 所以an+an-1>0 那么an-an-1=2 所以{an}等差 n=1时，a1^2+2a1=4S1=4a1 a1=2 (a1=0舍去)所以an=...
4Sn=An^2+2An 4Sn-1=An-1^2+2An-1 两式相减得 4An=An^2+2An-An-1^2-2An-1^2 移项得 (An+An-1)(An-An-1-2)=0 因为An>0 所以An=An-1+2 n=1时,A1=2 所以An=2+2(n-1)=2n 解析看不懂?免费查看同类题视频解析查看解答 ...
∴2Sn+1=an+12+n+1-4. 两式相减得2Sn+1-2Sn=an+12+n+1-4-(an2+n-4), 即2an+1=an+12-an2+1, 则an+12-2an+1+1=an2, 即(an+1-1)2=an2, ∴an+1-1=an,或an+1-1=-an, 即an+1-an=1或an+1+an=1 若an+1-an=1 即数列{an}为等差数列,公差d=1. 若an+1+an=1 ...
【解析】-|||-1-|||-1-|||--111-|||-n-4n2-1-2n-1八2n+1)-22n-12n+1-|||-Sn=a1+a2+….+an-|||-x1-+×3+…+22n-12n+1-|||-1-|||-1-1-|||-1-|||-1.11-|||-1-|||-1-|||-2-|||-×(1-3+3~5-|||-十…-|||-.+2n-1-2n+1-|||-1-|||-1-|||-×(...
解得a1=2,由4Sn=an2+2an4Sn=an2+2an,n用n-1代,两式相减得an2−an−12=2(an+an−1)an2−an−12=2(an+an−1),得an=2n.对n=1也成立.则数列{an}的通项公式为an=2n;(Ⅱ)bn=2n∙3nbn=2n•3n,错位相减法可以得Sn=2•3+4•32+…+2n•3n,3Sn=2•32+4•33+…...
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