数列{an满足a1=1,a2=2/3,且1/(an-1)+1/(an+1)=2/an,n≥2,n∈正整数,则an=( )

设1/an=bn则有2bn=(bn-1)+(bn+1)变形得(bn+1)-bn=bn-(bn-1)由此可以看出数列{bn}是等差数列公差d=(bn+1)-bn=bn-(bn-1)=````b2-b1=1/a2-1/a1=3/2-1=1/2b1=1/a1=1所以数列{bn}的通项公式为bn=b1+(n-1)d=1+(n-1)/2=(n+1...