已知数列{a n }的前n项和为S n , a 1 = 1 4 ,且2S n =2S n-1 +2a n-1 +1(n≥2,n∈N*)
已知数列{an}的前n项和为Sn,a1=14,且2Sn=2Sn-1+2an-1+1(n≥2,n∈N*).数列{bn}满足b1=34,且3bn-bn-1=n(n≥2,n∈N...
已知数列{a n }的前n项和为S n , a 1 = 1 4 ,且2S n =2S n-1 +2a n-1 +1(n≥2,n∈N*).数列{b n }满足 b 1 = 3 4 ,且3b n -b n-1 =n(n≥2,n∈N * ).(Ⅰ)求证:数列{a n }为等差数列;(Ⅱ)求证:数列{b n -a n }为等比数列;(Ⅲ)求数列{b n }的通项公式以及前n项和T n .
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人生如梦533
推荐于2016-12-06
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(Ⅰ)证明:∵2S n =2S n-1 +2a n-1 +1(n≥2,n∈N*), ∴当n≥2时,2a n =2a n-1 +1, 可得 a n - a n-1 = . ∴数列{a n }为等差数列.(4分) (Ⅱ)证明:∵{a n }为等差数列,公差 d= , ∴ a n = a 1 +(n-1)× = n- . (5分) 又3b n -b n-1 =n(n≥2), ∴ b n = b n-1 + n(n≥2) , ∴ b n - a n = b n-1 + n- n+ = b n-1 - n+ = ( b n-1 - n+ ) = [ b n-1 - (n-1)+ ] = ( b n-1 - a n-1 ). (8分) 又 b 1 - a 1 = ≠0 , ∴对n∈N*,b n -a n ≠0,得 = (n≥2) . ∴数列{b n -a n }是首项为 公比为 等比数列.(9分) (Ⅲ)由(Ⅱ)得 b n - a n = ?( ) n-1 , ∴ b n = - + ?( ) n-1 (n∈N*) .(11分) ∵ b 1 - a 1 + b 2 - a 2 ++ b n - a n = , ∴ b 1 + b 2 ++ b n -( a 1 + a 2 ++ a n )= [1-( ) n ] , ∴ T n - = [1-( ) n ] . ∴ T n = + [1-( ) n ] (n∈N*) .(14分) |
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