若数列{an}是公差为d的等差数列,则数列{an+

若数列{an}是公差为d的等差数列,则数列{an+2a(n+2)}是公差为多少的等差数列

1个回答

an = a1+(n-1)d
a(n+2)=a1+(n+1)d
an+2a(n+2)=a1+(n-1)d+2a1+2(n+1)d = 3a1+(3n-1)d = 3a1+(3n-3+2)d
=3a1+2d+(n-1)*3d
所以它的首项为3a1+2d,公差为3d