己知递增的等比数列{an}满足a₂+a₃+a₄=28且a₃+2是a₂,a₄的等差中项(1)求{an}的通项公式;(2)若bn=anlog(1/2)an,sn是数列{bn}的前n项和,求使sn+n.2^(n+1)>30成立的n的最小值.
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![己知递增的等比数列{an}满足a₂+a₃+a₄=28且a₃+2是a₂,a₄的等差中项(1)求{an}的通项公式;(2)若bn=anlog(1/2)an,sn是数列{bn}的前n项和,求使sn+n.2^(n+1)>30成立的n的最小值.](/uploads/image/z/8668576-64-6.jpg?t=%E5%B7%B1%E7%9F%A5%E9%80%92%E5%A2%9E%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a%26%238322%3B%2Ba%26%238323%3B%2Ba%26%238324%3B%3D28%E4%B8%94a%26%238323%3B%2B2%E6%98%AFa%26%238322%3B%2Ca%26%238324%3B%E7%9A%84%E7%AD%89%E5%B7%AE%E4%B8%AD%E9%A1%B9%281%29%E6%B1%82%EF%BD%9Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5bn%3Danlog%281%2F2%29an%2Csn%E6%98%AF%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E6%B1%82%E4%BD%BFsn%2Bn.2%5E%28n%2B1%29%3E30%E6%88%90%E7%AB%8B%E7%9A%84n%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC.)
己知递增的等比数列{an}满足a₂+a₃+a₄=28且a₃+2是a₂,a₄的等差中项(1)求{an}的通项公式;(2)若bn=anlog(1/2)an,sn是数列{bn}的前n项和,求使sn+n.2^(n+1)>30成立的n的最小值.
己知递增的等比数列{an}满足a₂+a₃+a₄=28且a₃+2是a₂,a₄的等差中项
(1)求{an}的通项公式;
(2)若bn=anlog(1/2)an,sn是数列{bn}的前n项和,求使sn+n.2^(n+1)>30成立的n的最小值.
己知递增的等比数列{an}满足a₂+a₃+a₄=28且a₃+2是a₂,a₄的等差中项(1)求{an}的通项公式;(2)若bn=anlog(1/2)an,sn是数列{bn}的前n项和,求使sn+n.2^(n+1)>30成立的n的最小值.
设公比为q.
a2+a3+a4=28 2(a3+2)+a3=28
3a3=24 a3=8
2(a3+2)=a2+a4
2a3+4=a3/q +a3q
a3=8代入,整理,得
2q²-5q+2=0
(2q-1)(q-2)=0
q=1/2(数列是递增数列,舍去)或q=2
an=a1q^(n-1)=a3q^(n-3)=8×2^(n-3)=2ⁿ
数列{an}的通项公式为an=2ⁿ
bn=anlog(1/2)(an)=2ⁿlog(1/2)(2ⁿ)=-n×2ⁿ
Sn=b1+b2+...+bn=-(1×2+2×2²+3×2³+...+n×2ⁿ)
2Sn=-[1×2²+2×2³+...+(n-1)×2ⁿ+n×2^(n+1)]
Sn-2Sn=-Sn=-[2+2²+...+2ⁿ-n×2^(n+1)]
Sn=2×(2ⁿ-1)/(2-1) -n×2^(n+1)
=(1-n)×2^(n+1)-2
Sn+n×2^(n+1)>30
(1-n)×2^(n+1)-2+n×2^(n+1)>30
2^(n+1)>32
2^(n+1)>2^5
n+1>5
n>4,n为正整数,n≥5,n的最小值是5.