在数列{a n }中,其前n项和S n 与a n 满足关系式:(t-1)S n +(2t+1)a n =t(t>0,n=1,2,3,…).
在数列{an}中,其前n项和Sn与an满足关系式:(t-1)Sn+(2t+1)an=t(t>0,n=1,2,3,…).(Ⅰ)求证:数列{an}是等比数列;(Ⅱ)设数列{a...
在数列{a n }中,其前n项和S n 与a n 满足关系式:(t-1)S n +(2t+1)a n =t(t>0,n=1,2,3,…).(Ⅰ)求证:数列{a n }是等比数列;(Ⅱ)设数列{a n }的公比为f(t),已知数列{b n }, b 1 =1, b n+1 =3f( 1 b n ) (n=1,2,3,…) ,求b 1 b 2 -b 2 b 3 +b 3 b 4 -b 4 b 5 +…+(-1) n+1 b n b n+1 的值.
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证明:(Ⅰ) 当n=1时,(t-1)S 1 +(2t+1)a 1 =t,∴a 1 =
当n≥2时,(t-1)S n +(2t+1)a n =t,(t-1)S n-1 +(2t+1)a n-1 =t ∴(t-1)a n +(2t+1)a n -(2t+1)a n-1 =0 ∴3ta n =(2t+1)a n-1 ,t>0 ∴
∴数列{a n }是以
(II)由(Ⅰ)可知, f(t)=
所以,数列{b n }是以2为公差,首项为1的等差数列 即b n =2n-1 ①当n为奇数时, b 1 b 2 -b 2 b 3 +b 3 b 4 -b 4 b 5 +…+(-1) n+1 b n b n+1 =b 1 b 2 +b 3 (b 4 -b 2 )+b 5 (b 6 -b 4 )+…+b n (b n+1 -b n-1 ) =3+4(b 3 +b 5 +…+b n ) =2n 2 +2n-1 ②当n为偶数时, b 1 b 2 -b 2 b 3 +b 3 b 4 -b 4 b 5 +…+(-1) n+1 b n b n+1 =b 2 (b 1 -b 3 )+b 4 (b 3 -b 5 )+…+b n (b n-1 -b n+1 ) =-4(b 2 +b 4 +…+b n ) =-(2n 2 +2n) 所以,原式=
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