已知数列an的前n项和为Sn

Question

如题所述

Answer 1

s(n+1)-s(n) + s(n+1)s(n)=0,
s(n)不为0。

1/s(n) - 1/s(n+1) + 1 = 0,
1/s(n+1) = 1/s(n) + 1,
{1/s(n)}是首项为1/s(1)=1/a(1)=1,公差为1的等差数列。
1/s(n) = 1 + (n-1) = n,
s(n) = 1/n.

s(n+1) = 1/(n+1).
a(n+1) = s(n+1)-s(n) = 1/(n+1) - 1/n = -1/[n(n+1)],

n=1时，a(n)=1,
n>=2时，a(n) = -1/[(n-1)n].

-1/20=a(m),所以，m>=2.

-1/20 = -1/(4*5) = a(m) = -1/[(m-1)m],
m=5.
a(5) = -1/20

答案：D。