…(12分)∴Tn=(n-1)•2n+1+2.(14分) (1)根据an+1=Sn+1-Sn,得到n≥2时an+1和an关系式即an+1=2an+1,两边同加1得到an+1+1=2(an+1),最后验证n=1时等式也成立,进而证明数列{an+1}是等比数列,并求数列{an}的通项公式;(2)利用错位相减法求数列{nan+n}的前n项和Tn. ...
2Sn=nAn 1式2S(n-1)=(n-1)A(n-1) 2式1式-2式得:2An =nAn -(n-1)A(n-1),整理得到:An/A(n-1)=(n-1)/(n-2)递推得到:A(n-1)/A(n-2)=(n-2)/(n-3).各式相乘得到An/A2=n-1由题干可知A2=1,所以An=n-1 (n>=2)... 分析总结。 各式相乘得到ana2n1由题干可知a21所以an...
2S(n-1)+a(n-1)=1 2Sn-2S(n-1)=2an=(1-an)-[1-a(n-1)]=a(n-1)-an 3an=a(n-1)an/a(n-1)=1/3 数列an为q=1/3等比数列
所以数列{an}是等差数列; (II)解:由(I)得数列{an}是等差数列,an=3n-2 由题意k≤ n (3n-2)•2n 存在正整数解 令bn= n (3n-2)•2n >0 因为 bn+1 bn = (n+1)(3n-2) 2(3n+1)n <1 所以{bn}为单调递减数列,故{bn}的最大值为b1= ...
数列{an}是以⅓为首项，⅓为公比的等比数列 an=⅓·⅓ⁿ⁻¹=⅓ⁿ数列{an}的通项公式为an=⅓ⁿ(2)bn=2a(n+1)/(1+an)[1+a(n+1)]=2·⅓ⁿ⁺¹/(1+⅓ⁿ)(1+⅓&#...
a1=1 2Sn=an+1/an 求通项公式 答案 2Sn=an+1/an ,an=Sn-Sn-1,所以2Sn=(Sn-Sn-1+1)\(Sn-Sn-1),整理Sn+Sn-1=1\(Sn-Sn-1),即Sn*Sn-Sn-1*Sn-1=1,{Sn平方}为等差数列,所以(Sn)^2=n,即Sn=√n,故an=√n-√(n-1).相关
当n=1时,2S1=a1,解得a1=0,当n≥2时,2Sn-1=(n-1)an-1,②,由①-②可得,2an=nan-(n-1)an-1,即(n-2)an=(n-1)an-1,∴(\;a_n)/(a_(n-1))=(n-1)/(n-2),∴(a_3)/(a_2)=2/1,(a_4)/(a_3)=3/2,…,(\;a_n)/(a_(n-1))=(n-1)/(n-2),...
∴n≥2时,2Sn=(n+1)an,两式相减,得:2an=(n+1)an-nan-1,nan-1=(n-1)an, an an−1= n n−1, an−1 an−2= n−1 n−2,… a3 a2= 3 2, a2 a1=2,∴an= a1• a2 a1× a3 a2× a4 a3×…× an an−1=a1× 2× 3 2× 4 3×…× n n−1=na1. 由已知...
(1)n=1时,a1=1.∵2Sn=3an-1,∴2Sn+1=3an+1-1,∴an+1=3an,∴an=3n-1.(2)∵bn=n⋅3n-1,∴Tn=1⋅30+2⋅31+3⋅32+…+(n-1)⋅3n-2+n⋅3n-1,3 Tn=1⋅31+2⋅32+3⋅33+…+(n-1)⋅3n-1+n⋅3n,两式相减可得-2Tn=1+31+32+…+3n-1-n⋅3n,∴Tn= 2n-1...
an an-1 =a1×2× 3 2 × 4 3 ×…× n n-1 ,由此能求出结果. 解答:解:∵2Sn=(n+1)an, ∴n≥2时,2Sn=(n+1)an, 两式相减,得:2an=(n+1)an-nan-1, nan-1=(n-1)an, an an-1 = n n-1 , an-1 an-2 = n-1