Question

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

Answer 1

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