已知数列{an}通项公式an=4n-2,且cn=(an^2+a(n+1)^2)/(2*an*a(n+1)),求Tn=c1+c2+..
问题描述:
已知数列{an}通项公式an=4n-2,且cn=(an^2+a(n+1)^2)/(2*an*a(n+1)),求Tn=c1+c2+..
答
先化简Cn
Cn=0.5an/a(n+1)+0.5a(n+1)/an=0.5[(2n-1)/(2n+1)+(2n+1)/(2n-1)]=1+1/(2n-1)-1/(2n+1)
Tn=n+(1-1/3)+(1/3-1/5)+(1/5-1/7)+……+[1/(2n-1)-1/(2n+1)]
=n+1-1/(2n+1)