等差数列公差为2,前n项和Sn=Pn方+2n若bn=﹙2n-1﹚An分之2,记数列{Bn}的前N项和为Tn,求使Tn﹥1
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等差数列公差为2,前n项和Sn=Pn方+2n若bn=﹙2n-1﹚An分之2,记数列{Bn}的前N项和为Tn,求使Tn﹥10分
之9的最小正整数n的值
之9的最小正整数n的值
![等差数列公差为2,前n项和Sn=Pn方+2n若bn=﹙2n-1﹚An分之2,记数列{Bn}的前N项和为Tn,求使Tn﹥1](/uploads/image/z/5306587-43-7.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E5%85%AC%E5%B7%AE%E4%B8%BA2%2C%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%3DPn%E6%96%B9%2B2n%E8%8B%A5bn%3D%EF%B9%992n-1%EF%B9%9AAn%E5%88%86%E4%B9%8B2%2C%E8%AE%B0%E6%95%B0%E5%88%97%EF%BD%9BBn%EF%BD%9D%E7%9A%84%E5%89%8DN%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E6%B1%82%E4%BD%BFTn%EF%B9%A51)
已知等差数列{an}的公差为2,其前n项和Sn=pn²+2n(n∈N*).
(I)求p的值及an;
(II)若bn=2/﹙2n-1﹚an,记数列{bn}的前n项和为Tn,求使Tn﹥9/10成立的最小正整数n的值
(I)∵{an}的等差数列
∴Sn=na1+n(n-1)d/2=na1+n(n-1)=n²+(a1-1)n
又由已知Sn=pn²+2n,
∴p=1,a1-1=2,
∴a1=3,
∴an=a1(n-1)d=2n+1
∴p=1,an=2n+1;
(II)由(I)知bn=2/(2n-1)(2n+1)=1/(2n-1)-1/(2n+1)
∴Tn=b1+b2+b3+…+bn
=(1-1/3)+(1/3-1/5)+(1/5-1/7)+……+[1/(2n-1)-1/(2n+1)]
=1-1/(2n+1)
=2n/(2n+1)
∵Tn>9/10
∴2n/(2n+1)>9/10,解得 n>9/2 又∵n∈N+
∴n=5
(I)求p的值及an;
(II)若bn=2/﹙2n-1﹚an,记数列{bn}的前n项和为Tn,求使Tn﹥9/10成立的最小正整数n的值
(I)∵{an}的等差数列
∴Sn=na1+n(n-1)d/2=na1+n(n-1)=n²+(a1-1)n
又由已知Sn=pn²+2n,
∴p=1,a1-1=2,
∴a1=3,
∴an=a1(n-1)d=2n+1
∴p=1,an=2n+1;
(II)由(I)知bn=2/(2n-1)(2n+1)=1/(2n-1)-1/(2n+1)
∴Tn=b1+b2+b3+…+bn
=(1-1/3)+(1/3-1/5)+(1/5-1/7)+……+[1/(2n-1)-1/(2n+1)]
=1-1/(2n+1)
=2n/(2n+1)
∵Tn>9/10
∴2n/(2n+1)>9/10,解得 n>9/2 又∵n∈N+
∴n=5
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