设an是由正数构成的等比数列,bn=a(n+1)+a(n+2),cn=an+a(n+3),则()a.bn>cn b.bn<cn c.bn≥cn d.bn≦cn
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![设an是由正数构成的等比数列,bn=a(n+1)+a(n+2),cn=an+a(n+3),则()a.bn>cn b.bn<cn c.bn≥cn d.bn≦cn](/uploads/image/z/14330534-14-4.jpg?t=%E8%AE%BEan%E6%98%AF%E7%94%B1%E6%AD%A3%E6%95%B0%E6%9E%84%E6%88%90%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2Cbn%3Da%28n%2B1%29%2Ba%EF%BC%88n%2B2%EF%BC%89%2Ccn%3Dan%2Ba%EF%BC%88n%2B3%EF%BC%89%2C%E5%88%99%EF%BC%88%EF%BC%89a.bn%EF%BC%9Ecn+b.bn%EF%BC%9Ccn+c.bn%E2%89%A5cn+d.bn%E2%89%A6cn)
设an是由正数构成的等比数列,bn=a(n+1)+a(n+2),cn=an+a(n+3),则()a.bn>cn b.bn<cn c.bn≥cn d.bn≦cn
设an是由正数构成的等比数列,bn=a(n+1)+a(n+2),cn=an+a(n+3),则()
a.bn>cn b.bn<cn c.bn≥cn d.bn≦cn
设an是由正数构成的等比数列,bn=a(n+1)+a(n+2),cn=an+a(n+3),则()a.bn>cn b.bn<cn c.bn≥cn d.bn≦cn
设an的公比为q,且q>0,an>0,则
bn=an(q+q^2)
cn=an(1+q^3)
cn-bn=an(q^3+1-q(q+1))=an(q+1)(q^2-q+1-q)=an(q+1)(q-1)^2>=0
所以,当q=1时,cn=bn,否则,cn>bn
因此,选D
选D
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