已知等差数列{an}的首项为a1=1，公差d不为0，等比数列{bn}满足b2=a2，b3=a5，b4=a14 已知等差数列{an}的首项a1=1，公差d＞0，等比数列{b...
\u5df2\u77e5\u7b49\u5dee\u6570\u5217{an}\u7684\u9996\u9879a1=1\uff0c\u516c\u5deed\uff1e0\uff0c\u7b49\u6bd4\u6570\u5217{bn}\uff0c\u6ee1\u8db3b2=a2\uff0cb3=a5\uff0cb4=a14\uff0e\uff081\uff09\u6c42\u6570\u5217{an}\u4e0e{bn\uff081\uff09\u7531\u9898\u610f\u5f97\uff081+4d\uff092=\uff081+d\uff09\uff081+13d\uff09\uff0cd\uff1e0\u89e3\u5f97d=2\u2026\uff083\u5206\uff09\u2234an=2n-1\u2026\uff084\u5206\uff09\u53c8b2=a2=3\uff0cb3=a5=9\uff0c\u6240\u4ee5{bn}\u7684\u516c\u6bd4\u4e3a3\uff0cbn=3n-1\u2026\uff086\u5206\uff09\uff082\uff09\u2235cn=2an-18=4n-20\u2026\uff087\u5206\uff09\u4ee4cn\u22640\u5f97n\u22645\u2026\uff089\u5206\uff09\u6240\u4ee5\u5f53n=4\u6216n=5\u65f6\uff0csn\u53d6\u6700\u5c0f\u503c-40\u2026\uff0812\u5206\uff09
\u5f88\u9ad8\u5174\u4e3a\u60a8\u89e3\u7b54\uff1a
1)
\u89e3\uff1a\u56e0\u4e3a\u7b49\u5dee\u6570\u5217{an}\u7684\u9996\u9879a1=1
\u6240\u4ee5a2=a1+d=1+d,a5=a1+4d=1+4d,a14=a1+13d=1+13d
\u56e0\u4e3a{bn}\u4e3a\u7b49\u6bd4\u6570\u5217
\u6240\u4ee5(b3)^2=b2*b4
\u53c8a2=b2,a5=b3,a14=b4
\u6240\u4ee5(a5)^2=a2*a14
\u5373(1+4d)^2=(1+d)*(1+13d)
\u6240\u4ee51+8d+16d^2=1+14d+13d^2
\u5373d^2-2d=0
\u6240\u4ee5d=2\u6216d=0
\u53c8\u56e0\u4e3ad\uff1e0
\u6240\u4ee5d=2
\u6240\u4ee5an=a1+(n-1)d=1+2(n-1)=2n-1
\u6240\u4ee5b2=a2=3,b3=a5=9
\u6545q=b3/b2=9/3=3
\u6240\u4ee5b1=b2/q=3/3=1
\u6240\u4ee5bn=b1*q^(n-1)=1*3^(n-1)=3^(n-1)
(2)
c1/b1+c2/b2+c3/b3+\u2026\u2026+Cn/bn=a(n+1)
c1/b1+c2/b2+c3/b3+\u2026\u2026+Cn/bn=2n
\u8bbecn/bn=gn
Tn=2n
gn=Tn-Tn-1=2 \uff08n n-1 \u4e3a\u4e0b\u6807\uff09
\u6240\u4ee5Cn/Bn=2
Cn=2*3^(n-1)
\u5219\uff0c\u8fd0\u7528\u7b49\u6bd4\u6570\u5217\u6c42\u548c\u516c\u5f0f
C1+C2+\u2026\u2026+C2011=2x\uff081-3^n\uff09\\uff081-3\uff09
=3^n-1
\u8c22\u8c22\uff0c\u795d\u4f60\u5f00\u5fc3\u3002\u6709\u5e2e\u52a9\u8bb0\u5f97\u91c7\u7eb3\u54e6
解：因为等差数列{an}的首项a1=1
所以a2=a1+d=1+d,a5=a1+4d=1+4d,a14=a1+13d=1+13d
因为{bn}为等比数列
所以(b3)^2=b2*b4
又a2=b2,a5=b3,a14=b4
所以(a5)^2=a2*a14
即(1+4d)^2=(1+d)*(1+13d)
所以1+8d+16d^2=1+14d+13d^2
即d^2-2d=0
所以d=2或d=0
又因为d＞0
所以d=2
所以an=a1+(n-1)d=1+2(n-1)=2n-1
所以b2=a2=3,b3=a5=9
故q=b3/b2=9/3=3
所以b1=b2/q=3/3=1
所以bn=b1*q^(n-1)=1*3^(n-1)=3^(n-1)
(2)
c1/b1+c2/b2+c3/b3+……+Cn/bn=a(n+1)
c1/b1+c2/b2+c3/b3+……+Cn/bn=2n
设cn/bn=gn
Tn=2n
gn=Tn-Tn-1=2
所以Cn/Bn=2
Cn=2*3^(n-1)
解：
,A2,A5,A14分别是等比数列<Bn>的B2,B3,B4
则，A5^2=A2A14即（1+4d)^2=（1+d)(1+13d)
解得，d=2；
则An=1+2（n-1）=2n-1；
B2=A2=3;B3=A5=9;B4=A14=27
所以Bn是以1为首项3为公比的等比数列
Bn=3^(n-1)
C1/B1+C2/B2+````+Cn/Bn=A(n+1)，则，C1/B1=A1.C1=3
C1/B1+C2/B2+````+Cn-1/Bn-1=An
两式相减Cn/Bn=2，则Cn=2x3^(n-1)（n>=2)
当n=1，不满足通式，则数列Cn是，除去第一项其余项成等比数列
则C1+C2+````+C2008=3+2x3+2x3^2+……+2x3^2007
=3+（2x3-2x3^2007x3）/(1-3)
= 3^2008
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