已知{An}为等差数列,Bn=A3n+1,求证数列Bn为等差数列.

问题描述:

已知{An}为等差数列,Bn=A3n+1,求证数列Bn为等差数列.

设An=an+d,a和d为常数,因为B(n+1)-Bn=A(3n+3)-A3n=a为一常数就得证啦

an=a+(n-1)d 所以bn=a3n+1=a+(3n-1)d +1 所以bn-b[n-1] =a+(3n-1)d +1 -[a+ (3n-3-1)d +1] =3d 所以bn数列也是等差数列