Question

已知函数f(x)=2x-1/2,若数列{an}满足a1=1,a(n+1)=f(an)+f(n),bn=an+2n+1

已知函数f(x)=2x-1/2,若数列{an}满足a1=1,a(n+1)=f(an)+f(n),bn=an+2n+1判断数列bn是等差数列还是等比数列

Answer 1

a1=1,所以a2=f(a1)+f(1)=2f(1)=2*(2-1/2)=3
b1=a1+2*1+1=4,b2=3+2*2+1=8
b(n+1)-bn=a(n+1)+2(n+1)+1-an-2n-1=an+2n+1=bn
所兄滑谨数以,b(n+1)=2bn,是等比数列羡晌腊

追问

是不可以用列举法推的哦！
由题意：
a(n+1)=2an-1/2+2n-1/2=2an+2n-1
故b(n+1)=a(n+1)+2(n+1)+1=2an+4n+2
bn=an+2n+1
则b(n+1)/bn=2
所以{bn}为等差数列