Question

已知各项均为正数的数列{a n }的前n项和为S n ,首项为a 1 ,且 ,a n ,S n 成等差数列.(1)求数列{a n }的

已知各项均为正数的数列{a n }的前n项和为S n ,首项为a 1 ,且 ,a n ,S n 成等差数列.(1)求数列{a n }的通项公式;(2)若 = ,设c n = ,求数列{c n }的前n项和T n .

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Answer 1

(1) a n =2 n-2       (2) T n =

解:(1)由题意知2a n =S n + ,a n >0, 当n=1时,2a 1 =a 1 + ,∴a 1 = . 当n≥2时,S n =2a n - , S n-1 =2a n-1 - , 两式相减得a n =2a n -2a n-1 , 整理得 =2, ∴数列{a n }是以 为首项,2为公比的等比数列. a n =a 1 ·2 n-1 = ×2 n-1 =2 n-2 . (2) = =2 2n-4 , ∴b n =4-2n, ∴c n = = , 即c n = . 则T n =c 1 +c 2 +c 3 +…+c n , 即T n = + + +…+ . ∴ T n = + + +…+ , 则 T n =4+ + +…+ - . T n =8-( + +…+ )+ =8- + =8-8(1- )+ = + = + = . 即T n = .