a12a23a3nan求an

卞爽会3948已知等比数列{an}中,a1=1,a4=27,设Tn=a1+2a2+3a3+…nan,求Tn的值 -
奚娄邦18090516661 ______ an=3^(n-1) tn=1+2*3+3*3^2+....+n*3^(n-1) 13tn=3+2*3^2+....(n-1)*3^(n-1)+n*3^n 22-1得2tn= -(1+3+3^2+...+3^(n-1))+n*3^n = -[1/2*(3^n-1)]+n*3^n tn=n/2*3^n-1/4*(3^n-1)

卞爽会3948已知数列{an}满足2a1+22a2+23a3+…+2nan=4n - 1,则{an}的通项公式------ -
奚娄邦18090516661 ______ ∵数列{an}满足2a1+22a2+23a3+…+2nan=4n-1,∴当n≥2时,2nan=(4n-1)-(4n-1-1),化为an=3?2n-2. 当n=1时,2a1=4-1,解得a1=3 2 ,上式也成立. ∴an=3?2n-2. 故答案为:an=3?2n-2.

卞爽会3948已知数列an满足2a1+22a2+…+2nan=4n - 1,则an=34?2n34?2n -
奚娄邦18090516661 ______ ∵2a1+22a2+…+2nan=4n-1…①,∴2a1+22a2+…+2n-1an-1=4n-1-1…②,①-②得2nan=3*4n-1,∴an=3 4 ?2n,故答案为:3 4 ?2n.

卞爽会3948已知数列{an}满足12a1+122a2+123a3+…+12nan=2n+1则{an}的通项公式an=6 (n=1)2n+1(n≥2)an=6 (n=1)2n+1(n≥2). -
奚娄邦18090516661 ______[答案] ∵数列{an}满足 1 2a1+ 1 22a2+ 1 23a3+…+ 1 2nan=2n+1,① ∴当n≥2时,仿仿写一个式子 1 2a1+ 1 22a2+…+ 1 2n−1an−1=2n−1② ①-②得 1 2nan=2, ∴an=2n+1n≥2, 当n=1时,a1=6, ∴{an}的通项公式 an= 6 (n=1)2n+1(n≥2) 故答案为:an= 6 ...

卞爽会3948已知数列{An}满足A1=2╱3,An+1=(n╱(n+1))An,求An:(详解,谢谢) -
奚娄邦18090516661 ______ (n+1)A(n+1) =nAn A(n+1)=n(An‐A(n+1) A2=A1‐A2 A3=2(A2‐A3) A4=3(A3‐A4) .....A2+A3+A4+....A(n+1)=A1+A2+A3+...‐nA(n+1) A(N+1)=A1/﹙N+1﹚=2/﹙3N+3﹚ 满意请及时采纳!

卞爽会3948已知数列{an}满足a1=2,an+1=2n+1an(n+12)an+2n,n∈N*(1)设bn=2nan,求数列bn的通项公式.(2)设c -
奚娄邦18090516661 ______ (1)由bn=2n an ,bn+1=2n+1 an+1 ,得到an=2n bn ,an+1=2n+1 bn+1 ,b1=2 a1 =1. 代入an+1=2n+1an (n+1 2 )an+2n ,化为bn+1?bn=n+1 2 . ∴bn=(bn-bn-1)+(bn-1-bn-2)+…+(b2-b1)+b1=(n-1)+1 2 +(n-2)+1 2 +…+1+1 2 +1= n(n?1) 2 + n?1 2 +1= ...

卞爽会3948已知数列{an}为等差数列,前n项和为Sn,且S3=12,a3份2=a1+a2,若a3<0,求数列{an}的通项公式及前n项和QSn -
奚娄邦18090516661 ______ a(n)=a+(n-1)d12=s(3)=3a+3d, a=4-d [a(3)]^2 = [a+2d]^2=a(1)+a(2)=2a+d0=d^2+9d+8=(d+1)(d+8) d=-1或d=-8.若d=-1,a=5,则a(3)=a+2d=3>0与a(3)<0矛盾.因此,d=-8,a=12 a(n)=12-8(n-1),s(n)=12a-4n(n-1)

卞爽会3948已知数列{an}满足:1•a1+2•a2+3•a3+…n•an=n(1)求{an}的通项公式;(2)若bn=2nan,求{bn}的前n项和Sn. -
奚娄邦18090516661 ______[答案] (1)∵数列{an}满足:1•a1+2•a2+3•a3+…n•an=n,∴当n≥2时,nan=ni=1i•ai-n−1ii•ai=1,∴an=1n,当n=1时,a1=1成立,∴an=1n.(2)∵bn=n•2n,∴Sn=1•21+2•22+3•23…+n•2n①2Sn=1•22+2•23+3...

卞爽会3948(2013•潮州二模)已知各项都不为零的数列{an}的前n项和为Sn,且Sn=12anan+1(n∈N*),a1=1.(1)求数列{an}的通项公式;(2)求证:1a12+1a22+1a32+…... -
奚娄邦18090516661 ______[答案] (1)∵Sn=12anan+1,①∴Sn−1=12an−1an(n≥2),②①-②得an=Sn−Sn−1=12(an+1−an−1)an∵an≠0,∴an+1-an-1=2.数列{an}的奇数项组成首项为a1=1,公差为2的等差数列;偶数项组成首项为a2,公差为2的等差数...

卞爽会3948(2010•黄浦区一模)(理科)已知各项都为正数的数列{an}满足a1=1,Sn=12anan+1(n∈N+),其中Sn是数列{an}的前n项的和.(1)求数列{an}的通项公式an;... -
奚娄邦18090516661 ______[答案] (1)∵sn=12an•an+1,(n∈N*),∴sn−1=12an−1•an.∴an=12an(an+1-an-1),即an+1-an-1=2(n≥2).∴a2,a4,a6,…a2n是首项为a2,公差为2的等差数列; a1,a3,…a2n-1是首项为a1,公差为2的等差...