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

4Sn =an +1
By MI : an = - (-1/3)^n
n=1 , a1= 1/3
p(1) is true
Assume p(k) is true
ak= -(-1/3)^k
for n=k+1
a(k+1) = S(k+1) - Sk
= (1/4)( a(k+1) - ak)
3a(k+1) = -ak
= (-1/3)^k
a(k+1) = -(-1/3)^(k+1)
p(k+1) is true
By principle of MI, it is true for all n