Question

设各项均为正数的无穷数列an和bn满足2bn=an+an+1且an-1方=bn*bn+1,求证根号bn是等差数列

a1=1,a2=2求an和bn的通项公式

Answer 1

题意： an+an+1=2bn; (1)
bnbn+1=an+1*an+1 (2)
（2）式两边开方得：an+1=sqrt(bn)*sqrt(bn+1) (3)
(1)式两液饥边平方，展开,然后将槐卖（3）式代入，可得：
bn*bn-1+bn*bn+1+2*sqrt(bn-1*bn*bn*bn+1)=4bn*bn (4)
整理闹明返（4）式后可得
sqrt(bn-1)+sqrt(bn+1)=2sqrt(bn)