解:(1)设等比数列{an}的公比为q>1,∵a2=6,a1+a2+a3=26, ∴6 +6+6g=26, 化为3q2-10q+3=0,q>1. 解得q=3, ∴an=19=6×3n-2=2×3n-1. (2)设等差数列{cn}的公差为d, cn=an+bn,b1=a1, ∴c1=2a1=4. c3=a3+b3=18-10=8, ∴8=4+2d,解得d=2. ∴cn=4+2(n-1)=2n...
化为3q2-10q+3=0,q>1.解得q=3,∴an=a2qn-2=6×3n-2=2×3n-1.(2)设等差数列{cn}的公差为d,cn=an+bn,b1=a1,∴c1=2a1=4.c3=a3+b3=18-10=8,∴8=4+2d,解得d=2.∴cn=4+2(n-1)=2n+2.∴bn=cn-an=2(n+1)-2×3n-1....
Compare AT25040AN-10SQ-2.7SL383 by undefined vs ST24W04B3 by undefined. View differences in part data attributes and features.
已知数列{an}是等差数列,数列{bn}是正项等比数列,且满足a1=1,b1=4,a2+b2=10,a26-b3=10.(1)求数列{an},{bn}的通项公式;(2)记cn=anbn,求数列{Cn}的前n项和Sn.
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已知Sn为等差数列{bn}的前n项和,且满足3b2=b5,b3=5b2-10,若数列{an}满足an an 1=bn,b1=a1 1,则( ) A. b32=63 B
解:(1)设等差数列{an}的公差为d,∵a1+a2=10,a5-a3=4.∴2a1+d=10,2d=4,联立解得a1=4,d=2,an=4+2(n-1)=2n+2.(2)设等比数列{bn}的公比为q,由b2=a3=8,b3=a7=16=qb2,解得q=2.∴2b1=8,解得b1=4,∴b4=4×23=32=2n+2,解得n=15.∴b4是数列{an}的第15项. (1)设等差...
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an 2 (an≥2).当am为奇数时,必然有 an-1=an-1,(an≥2),an-1是偶数,可继续重复上面的操作. 所以要使项数m最小,只需遇到偶数除以2,遇到奇数则减1.由此可得 m=b1+(b2-b1)+(b3-b2)+(b4-b3)+…+(bl-bl-1)+(l-1)+1=bl+l. 解答:解:(Ⅰ)1,2,3,4,5,10或1,2,4,8,9,10. …(2...
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