Question

急 高中数学 通项公式an=(1+2+……+n)/n, bn=1/(an×an+1） 求bn前n项的和，注最后n+1是下标

Answer 1

an=（1+n)/2
bn=4/(1+n)(2+n)=4[1/(1+n)-1/(2+n)]
Tn=b1+b2+...+bn
=4[(1/2-1/3)+(1/3-1/4)...+(1/(n+1)-1/(n+2)]
=4[1/2-1/(n+2)]
=2-4/(n+2)

Answer 2

解：
an=(1+2+...+n)/n=n(n+1)/(2n)=(n+1)/2
bn=1/[ana(n+1)]=4/[(n+1)(n+2)]=4/(n+1)-4/(n+2)
Sn=b1+b2+...+bn
=4/2-4/3+4/3-4/4+...+4/(n+1)-4/(n+2)
=4/2-4/(n+2)
=2n/(n+2)

Answer 3

An = (1+2+...+n)/n
= (n(n+1)/2) / n
= (n+1)/2

所以Bn = 1/(An * A(n+1))
= 1/ ( (n+1)/2 * (n+2)/2)
= 4 / (n+1)(n+2)
= 4/(n+1) - 4/(n+2)

因此Bn的前n项和Sn = B1 + B2 + ... + Bn
= (4/2 - 4/3) + (4/3 - 4/4) + ... + (4/(n+1) - 4/(n+2))
= 4/2 - 4/(n+2)
= 2 - 4/(n+2)
= 2n/(n+2)