已知等差数列{an}的各项均为正数,an>0,a1=1,前n项和为Sn,{bn}为等比数列,b1=1,b2s2=6,b3
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已知等差数列{an}的各项均为正数,an>0,a1=1,前n项和为Sn,{bn}为等比数列,b1=1,b2s2=6,b3s3等于24,n∈N*
第一问:求an,bn
第二问:令Cn=(n/bn)+(1/an*an+2),Tn为Cn的前n项和,求Tn
第三问:记F(k)=19/2 - 2Tk - k+2/2^k-2 (k∈N*),若Fk≥21/110恒成立,求Kmax
需要详细过程,谢谢
第一问:求an,bn
第二问:令Cn=(n/bn)+(1/an*an+2),Tn为Cn的前n项和,求Tn
第三问:记F(k)=19/2 - 2Tk - k+2/2^k-2 (k∈N*),若Fk≥21/110恒成立,求Kmax
需要详细过程,谢谢
![已知等差数列{an}的各项均为正数,an>0,a1=1,前n项和为Sn,{bn}为等比数列,b1=1,b2s2=6,b3](/uploads/image/z/7234738-34-8.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%2Can%3E0%2Ca1%3D1%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%7Bbn%7D%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2Cb1%3D1%2Cb2s2%3D6%2Cb3)
(1)设公差为d,公比为q
则b2S2=q(2+d)=6
b3S3=q^2(3+3d)=24
解得d=1或d=-1/2(舍去)
q=2
所以an=n,bn=2^(n-1)
(2)cn=n/2^(n-1)+1/[n(n+2)]
这两项的求和分别用错位相减法和裂项相加法
{n/2^(n-1)}的前n项和(设为P):
P=1/1+2/2+3/2^2+...+(n-1)/2^(n-2)+n/2^(n-1)
1/2P=1/2+2/2^2+2/2^3+...+(n-1)/2^(n-1)+n/2^n
两式相减,得1/2P=1+1/2+1/2^2+...+1/2^(n-1)-n/2^n=2-1/2^(n-1)-n/2^n=2-(n+2)/2^n
P=4-(n+2)/2^(n-1)
{1/[n(n+2)]}的前n项和(设为Q):
Q=1/(1*3)+1/(2*4)+...+1/[n(n+2)]=1/2(1-1/3)+1/2(1/2-1/4)+...+1/2(1/n-1/(n+2))
=1/2(1+1/2-1/(n+1)-1/(n+2))=3/4-(2n+3)/[2(n+1)(n+2)]
Tn=19/4-(n+2)/2^(n-1)-(2n+3)/[2(n+1)(n+2)]
(3)F(k)=(2n+3)/[(n+1)(n+2)]=1/(k+1)+1/(k+2)
得知F(k)为递减数列
令1/(k+1)+1/(k+2)=21/110
得k=9
所以k(max)=9
则b2S2=q(2+d)=6
b3S3=q^2(3+3d)=24
解得d=1或d=-1/2(舍去)
q=2
所以an=n,bn=2^(n-1)
(2)cn=n/2^(n-1)+1/[n(n+2)]
这两项的求和分别用错位相减法和裂项相加法
{n/2^(n-1)}的前n项和(设为P):
P=1/1+2/2+3/2^2+...+(n-1)/2^(n-2)+n/2^(n-1)
1/2P=1/2+2/2^2+2/2^3+...+(n-1)/2^(n-1)+n/2^n
两式相减,得1/2P=1+1/2+1/2^2+...+1/2^(n-1)-n/2^n=2-1/2^(n-1)-n/2^n=2-(n+2)/2^n
P=4-(n+2)/2^(n-1)
{1/[n(n+2)]}的前n项和(设为Q):
Q=1/(1*3)+1/(2*4)+...+1/[n(n+2)]=1/2(1-1/3)+1/2(1/2-1/4)+...+1/2(1/n-1/(n+2))
=1/2(1+1/2-1/(n+1)-1/(n+2))=3/4-(2n+3)/[2(n+1)(n+2)]
Tn=19/4-(n+2)/2^(n-1)-(2n+3)/[2(n+1)(n+2)]
(3)F(k)=(2n+3)/[(n+1)(n+2)]=1/(k+1)+1/(k+2)
得知F(k)为递减数列
令1/(k+1)+1/(k+2)=21/110
得k=9
所以k(max)=9
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