强人进

题目

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?

答案

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)