Question

已知数列{an}的前n项和Sn，满足an+Sn=2n（1）证明数列{an -2)为等比数列，并求an

Answer 1

an=sn-s（n-1）= 2n-an-2n+2+a（n-1）= (a（n-1）+2 )/2
a1=1
[an-2]：[a（n-1）-2 ] = { [a(n-1)+2]/2 - 2 }/[a(n-1)-2] = 1/2 所以{an-2}等比。
{an-2}={-1，-1/2, -1/4, -1/8, ...}
an=2-1/(2^(n-1))

Answer 2

a_n + S_n = 2n
a_(n-1) + S_(n-1) = 2(n-1)

两式相减 得到
a_n - a_(n-1) + a_n = 2 =>

a_n - 2 = 1/2 * (a_(n-1) - 2)

所以an- 2是公比为1/2的等比数列
故有 a_n = (a_0- 2) / 2^n + 2

Answer 3

由已知得a(n-1)+s(n-1)=2*(n-1)。故an-a(n-1)+sn-s(n-1)=an-a(n-1)+an=2.即an=(1/2)*a(n-1)+1。所以an-2=(1/2)*[a(n-1)-2] 即[an-2]/[a(n-1)-2]=1/2。得数列{an -2)为等比数列公比为1/2，a1=1得：an=(1/2)^(n-1)+2