金典小魔女
幼苗
共回答了14个问题采纳率:92.9% 举报
(1)因为a 1 =1,则a 2 =1+d,a 5 =1+4d,a 14 =1+13d,
又等差数列{a n }的第2项、第5项、第14项分别为等比数列{b n }的第2项、第3项、第4项,
∴(1+4d) 2 =(1+d)(1+13d),
即3d(d-2)=0,又公差d>0,∴d=2,
则a n =a 1 +(n-1)d=1+2(n-1)=2n-1.
又b 2 =a 2 =3,b 3 =a 5 =9,
∴数列{b n }的公比为3,
则b n = b 2 q n-2 =3•3 n-2 =3 n-1 .
(2)由
c 1
b 1 +
c 2
b 2 +…+
c n
b n = a n+1 ①
当n=1时,
c 1
b 1 =a 2 =3,∴c 1 =3,
当n>1时,
c 1
b 1 +
c 2
b 2 +…+
c n-1
b n-1 =a n②
①-②得
c n
b n =a n+1 -a n =2(n+1)-1-(2n-1)=2
∴c n =2b n =2•3 n-1 (n>1),而c 1 =3不适用该通项公式.
∴c n =
3n=1
2• 3 n-1 n≥2 .
∴c 1 +c 2 +c 3 +…c 2012 =3+2•3+2•3 2 +…+2•3 2011
=1+2•1+2•3+2•3 2 +…+2•3 2011 =1+2•
1- 3 2012
1-3 =3 2012 .
1年前
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