数列﹛an﹜满足an=2a（n+1）-1,怎么证明等差?

数列﹛an﹜满足an=2a（n+1）-1,怎么证明等差? 数列[An]满足a1=2,a（n+1）=3an-2 求an 数列{an}满足a1=2,a(n+1)=2an+n+2,求an 已知数列an满足条件a1=-2 a（n+1）=2an/(1-an) 则an= 已知数列an满足a1=2,an=a(n-1)+2n,（n≥2）,求an 已知数列{an}满足a1=33,a（n+1）-an=2n,求an/n的最小值 已知数列{an}满足a1=2,a(n+1)-an=a(n+1）*an,则a31=? 已知数列｛An｝满足a1=1,a（n+1）=2an+1 求证数列{an+1}是等比数列 求数列｛an｝通式 在数列an中,a1=1,且满足a（n+1）=3an +2n,求an 数列an满足a1=1,a(n+1)=an/[(2an)+1],求a2010 已知数列｛an｝满足：a1=1,且an-a（n-1）=2n.求a2,a3,a4.求数列｛an｝通项an 已知数列{an}满足2an/an+2=an+1(n属于正整数),a1=1/1006.求证:数列{1/an}是等差数列,并求通项an已知数列{an}满足（2an）/（an+2）=a（n+1）(n属于正整数),a1=1/1006.求证:数列{1/an}是等差数列,并求通项an， 周期性数列问题i已知数列｛an}满足a(n+1)=2an (0 已知数列﹛an﹜满足a1=1,1/a(n+1)=√1/an²+2,an>0,求an 已知数列{a}满足a1=1/2,a（n+1）=an+1/(n^2+n),求an已知数列{a}满足a1=1/2,a（n+1）=an+1/(n^2+n),求an 已知数列an满足1/a-an=2根号n,且an>0.求an的通项公式是数列{an}满足1/an-an=2根号n，且an>0，求an的通项公式。 数列{an}满足a1=1 an+1=2n+1an/an+2n 数列an满足,a（n+1）=1+1/(1+an),证明an的极限是根号2