正数数列{an}的前n项和Sn,满足2√Sn=a(n+1),猜测an的表达式并证明

题目

正数数列{an}的前n项和Sn,满足2√Sn=a(n+1),猜测an的表达式并证明
请用数学归纳法证明,表示做到n=k+1时,a(k+1)=S(k+1)-Sk然后该怎么做?
我发现打错题目了- -是an + 不是a(n+1)

答案

4Sn =an +1By MI : an = - (-1/3)^nn=1 , a1= 1/3p(1) is trueAssume p(k) is trueak= -(-1/3)^kfor n=k+1a(k+1) = S(k+1) - Sk = (1/4)( a(k+1) - ak)3a(k+1) = -ak = (-1/3)^ka(k+1) = -(...