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...
(2)求和:b1+b3+b5+…+b2n-1. 相关知识点: 试题来源: 解析 【答案】(1);(2) 【解析】试题分析:(1)第(1)问,一般先求数列{an}的基本量,再求数列的通项.(2)第(2)问,先求数列{bn}的通项,再利用等比数列的求和公式求和.试题解析:(1)设等差数列{an}的公差为d.因为a2+a4=10,所以2a1...
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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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...
解:(I)设等差数列{an}的公差为d.∵a4-a3=2,所以d=2∵a1+a2=10,所以2a1+d=10∴a1=4,∴an=4+2(n-1)=2n+2(n=1,2,…)(II)设等比数列{bn}的公比为q,∵b2=a3=8,b3=a7=16,∴\((array)l(b_1q=8)(b_1q^2=16)(array).∴q=2,b1=4∴b_6=4*2^(6-1)=128,而128=2n+2...
从而b1+b3+b5+…+b2n-1=1+3+32+…+3n-1=. 相关知识点: 试题来源: 解析 (1) an=2n−1 (2) \frac{3^n - 1}{2} (1) 设等差数列{aₙ}的公差为d。 由a₁=1,且a₂ + a₄=10,得: (1+d) + (1+3d) = 10 ⇒ 2 + 4d = 10 ⇒ d=2。 故通项公式为aₙ =...
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(Ⅰ)等差数列{an},a1=1,a2+a4=10,可得:1+d+1+3d=10,解得d=2,所以{an}的通项公式:an=1+(n-1)×2=2n-1.(Ⅱ)由(Ⅰ)可得a5=a1+4d=9,等比数列{bn}满足b1=1,b2b4=9.可得b3=3,或-3(舍去)(等比数列奇... 解析看不懂?免费查看同类题视频解析查看解答 特别推荐 热点考点 2022年高考真题...