Question

数列{an}满足a1+2a2+2^2a3+```````+2^(n-1)an=n^2(n属于整数)(1)求{an}的通项

Answer 1

(1). a1=1^2=1

a1+2a2+2^2a3+```````+2^(n-2)a(n-1)+2^(n-1)an=n^2,

a1+2a2+2^2a3+```````+2^(n-2)a(n-1)=(n-1)^2,相减得:

an=(n^2-(n-1)^2)/2^(n-1)=(2n-1)/2^(n-1),n>1

{an}的通项公式:

an=(2n-1)/2^(n-1)

(2)..?