强人进1.设Sn=1/2+1/6+1/8+.+1/n(n+1),且,Sn*Sn+1=3/4,求n的值(过程)2.已知数列an=(1+2+3+.+n)/n,bn=1/(an*an+1),则bn前n项和为多少?
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![强人进1.设Sn=1/2+1/6+1/8+.+1/n(n+1),且,Sn*Sn+1=3/4,求n的值(过程)2.已知数列an=(1+2+3+.+n)/n,bn=1/(an*an+1),则bn前n项和为多少?](/uploads/image/z/5455434-66-4.jpg?t=%E5%BC%BA%E4%BA%BA%E8%BF%9B1.%E8%AE%BESn%3D1%2F2%2B1%2F6%2B1%2F8%2B.%2B1%2Fn%28n%2B1%29%2C%E4%B8%94%2CSn%2ASn%2B1%3D3%2F4%2C%E6%B1%82n%E7%9A%84%E5%80%BC%EF%BC%88%E8%BF%87%E7%A8%8B%EF%BC%892.%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%3D%EF%BC%881%2B2%2B3%2B.%2Bn%EF%BC%89%2Fn%2Cbn%3D1%2F%EF%BC%88an%2Aan%2B1%EF%BC%89%2C%E5%88%99bn%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BA%E5%A4%9A%E5%B0%91%3F)
强人进1.设Sn=1/2+1/6+1/8+.+1/n(n+1),且,Sn*Sn+1=3/4,求n的值(过程)2.已知数列an=(1+2+3+.+n)/n,bn=1/(an*an+1),则bn前n项和为多少?
强人进
1.设Sn=1/2+1/6+1/8+.+1/n(n+1),且,Sn*Sn+1=3/4,求n的值(过程)
2.已知数列an=(1+2+3+.+n)/n,bn=1/(an*an+1),则bn前n项和为多少?
强人进1.设Sn=1/2+1/6+1/8+.+1/n(n+1),且,Sn*Sn+1=3/4,求n的值(过程)2.已知数列an=(1+2+3+.+n)/n,bn=1/(an*an+1),则bn前n项和为多少?
1.
Sn=1/2+1/6+1/12+…+1/n(n+1)
=1/(1*2)+1/(2*3)+...+1/(n)*(n+1)
=1-1/2+1/2-1/3+...+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
所以
Sn+1=1-1/(n+2)=(n+1)/(n+2)
Sn*S(n+1)=3/4
n/(n+1)*(n+1)/(n+2)=3/4
n/(n+2)=3/4
4n=3n+6
n=6
2.
an
=(1+2+3+...+n)/n
=[(1+n)n/2]/n
=(1+n)/2
则:
a(n+1)
=(n+2)/2
则:
bn=1/[an*a(n+1)]
=1/[(n+1)/2]*[(n+2)/2]
=4/[(n+1)(n+2)]
则bn前n项和:
Sn=b1+b2+b3+...+bn
=4/[2*3]+4/[3*4]+4/[4*5]+...+4/[(n+1)(n+2)]
=4*[1/(2*3)+1/(3*4)+1/(4*5)+..+1/(n+1)(n+2)
=4*[(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+...+1/(n+1)-1/(n+2)]
=4*[1/2-1/(n+2)]
=2-4/(n+2)
=2n/(n+2)
这个是高一调研考试的原题。。。
1.
Sn=1/2+1/6+1/12+…+1/n(n+1)
=1/(1*2)+1/(2*3)+...+1/(n)*(n+1)
=1-1/2+1/2-1/3+...+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
所以
Sn+1=1-1/(n+2)=(n+1)/(n+2)
Sn*S(n+1)=3/4
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1.
Sn=1/2+1/6+1/12+…+1/n(n+1)
=1/(1*2)+1/(2*3)+...+1/(n)*(n+1)
=1-1/2+1/2-1/3+...+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
所以
Sn+1=1-1/(n+2)=(n+1)/(n+2)
Sn*S(n+1)=3/4
n/(n+1)*(n+1)/(n+2)=3/4
n/(n+2)=3/4
4n=3n+6
n=6
2.
an
=(1+2+3+...+n)/n
=[(1+n)n/2]/n
=(1+n)/2
则:
a(n+1)
=(n+2)/2
则:
bn=1/[an*a(n+1)]
=1/[(n+1)/2]*[(n+2)/2]
=4/[(n+1)(n+2)]
则bn前n项和:
Sn=b1+b2+b3+...+bn
=4/[2*3]+4/[3*4]+4/[4*5]+...+4/[(n+1)(n+2)]
=4*[1/(2*3)+1/(3*4)+1/(4*5)+..+1/(n+1)(n+2)
=4*[(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+...+1/(n+1)-1/(n+2)]
=4*[1/2-1/(n+2)]
=2-4/(n+2)
=2n/(n+2) 有需要我会帮你
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