Question

已知数列{an}的前N项和为Sn=n^2+2n(n属于N*），已求{an}通项公式为an=2n+1,

设Tn=1/a1*a2 +1/a2*a3 +1/a3*a4 +...+1/an*a(n+1) [n+1为下标],求Tn的值。

Answer 1

∵an=2n+1
∴a(n+1)-an=[2(n+1)+1]-(2n+1)=2

∴Tn=1/a1*a2 +1/a2*a3 +1/a3*a4 +...+1/an*a(n+1)
=(1/2)[(1/a1-1/a2)+(1/a2-1/a3)+(1/a3-1/a4)+...+(1/an-1/a(n+1))]
=(1/2)[1/a1-1/a(n+1)]
=(1/2)[1/(2*1+1)-1/(2(n+1)+1)]
=(1/2)[1/3-1/(2n+3)]
=n/(6n+9)

Answer 2

Tn=1/a1*a2 +1/a2*a3 +1/a3*a4 +...+1/an*a(n+1)=(a2-a1)/2/a1*a2+(a3-a2)/2/a2*a3+(a4-a3)/2/a3*a4+...+(a(n+1)-an)/2/an*a(n+1)=1/2/(1/a1-1/a2+1/a2-1/a3+1/a3-1/a4+...+1/an-1/a(n+1))=1/2(1/a1-1/a(n+1))=1/3/(2n+3)