Question

数列{an}中,a1=1sn是{an}的前n项和,当n大于等于2是sn=an[1-2/sn]求证{1/sn}是等差数列

Answer 1

1/Sn=(n+1)/2 则Sn=2/(n+1)

S1xS2+S2xS3+.+SnxS(n+1)=4(1/2 x 1/3 +1/3 x 1/4 +.)

=4(1/2 - 1/3 +1/3 -1/4 +.)

=4(1/2 -1/(n+2))