(1)∵a 2 =1+d,a 5 =1+4d,a 14 =1+13d,且a 2 ,a 5 ,a 14 成等比数列, ∴(1+4d) 2 =(1+d)(1+13d),解得d=2, ∴a n =1+(n-1)?2=2n-1, 又b 1 =a 2 =3,b 2 =a 5 =9, ∴q=3, b n =3? 3 n-1 = 3 n ; (2)
则n≥2时,
①-②得,
n=1时,C 1 =9, 所以 C n =
所以c 1 +c 2 +…+c 2013 =9+2?3 2 +2?3 3 +…+2?3 2013 =9+2?
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