是.因为数列{a n+1 -an}是等差数列,bn=a n+1 -an,所以{bn}是等差数列.设{bn}的公差为d.所以 b[3(n+1)-2]-b(3n-2)=b(3n+1)-b(3n-2)=[b(3n+1)-b3n]+[b3n-b(3n-1)]+[b(3n-1)-b(3n-2)]=3d因此{b3n-2}是等差数列...