(1)设等差数列的公差为d,则 ∵S 3 +S 5 =50,a 1 ,a 4 ,a 13 成等比数列, ∴3a 1 +3d+5a 1 +10d=50,(a 1 +3d) 2 =a 1 (a 1 +12d) ∵公差d≠0,∴a 1 =3,d=2 ∴数列{a n }的通项公式a n =2n+1; (2)据题意得b n = a 2 n =2×2 n +1. ∴数列{b n }的前n项和公式:T n =(2×2+1)+(2×2 2 +1)+…+(2×2 n +1)=2×(2+2 2 +…+2 n )+n=2×
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