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已知数列An为等差数列

Sn有最大值,说明d0.a11/a10 a11 0,|a11| > |a10|.a11 = a1 + 10d 2a1 + 20d a10 = a1 + 9d > 0.=>2a1 + 18d > 0=>a10 + a11 = 2a1 + 19d Sn = n*a1 + n(n-1)*d/2 =>a1 + (n-1)*d/2 =>2a1 + (n-1)*d

(1)∵数列{an}为等差数列,且满足a2=3,a4+a5+a6=18,设公差为d,∴a1+d=3a1+3d+a1+4d+a1+5d=18,解得a1=2,d=1,∴an=n+1.∵数列{bn}满足b1=1,bn+1=2bn+1,∴bn+1+1=2(bn+1),b1+1=2,∴{bn+1}是首项为2,

法1:因为数列{an}为等差数列,它的前n项和Sn可以写成Sn=An^2+Bn的形式,不含常数项,因为Sn=(n+1)^2+λ=n^2+2n+1+λ所以1+λ=0λ=-1法2:根据Sn求a1,a2,a3,由a1,a2,a3成等差数列,则2a2=a1+a3可求λ.法3:由Sn可先求数列{an}的通项公式,由于数列{an}为等差数列

(1)设等差数列{an}的公差为d,等比数列{bn}的公比为q,∵a2=2,a5+a9=14,∴a1+d=2,2a1+12d=14,解得a1=d=1.∴an=1+(n-1)=n.∴b1=a2=2,b4=a15+1=16=2*q3,∴q=2.∴bn=2n.(2)cn=anbn=n2n.∴数列{cn

设a2=k,a1=k-a,a3=k+a b1*b3=b2*b2 解出来k=-a/3 b1=-a/3 b2=2a/3 b3=-4a/3 公比q=-2

因为a4=a1+3d=2a9=a1+8d=-6所以解得a1=31/8,d=-5/8所以s10=10*(31/8)+[(10*9)/2]*(-5/8)=85/8望采纳

a1-a4-a8-a12+a15=2a1+a15-(a4+a12)-a8=2-a8=2a8=-2a3+a13=2a8=-4S15=15*(a1+a15)/2=15*2*a8/2=-30

a1+a7=20a1+a1+6d=20a1+3d=10a11-a8=18.3d=18d=6a1=-8an=6d-14

①设公差为d,公比为q∵数列{an+bn}的前三项依次为3,7,13∴a1+b1=3a2+b2=7a3+b3=13又a1=1∴b1=2d=2q=2∴an=2n-1,bn=2n②∵an=2n-1,bn=2n∴an+bn=(2n-1)+2n∴Sn=(a1+a2+…+an)+(b1+b2+…+bn)=(1+2n

由等差数列的性质得a4+a5+a6=3a5=π/4 则a5=π/12 故a1+a9=2a5=π/6则S9=9(a1+a9)/2=3π/4 故cosS9=-2算数平方根/2

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