Question

设数列{an}的前n项和为sn，已知sn=2an-2^(n+1)， 求证数列{an/2^n}为等差数列,并求{an}的通项公式

Answer 1

s(n) = 2a(n) - 2^(n+1),
a(1) = s(1) = 2a(1) - 2^2, a(1) = 4.

s(n+1) = 2a(n+1) - 2^(n+2),
a(n+1) = s(n+1)-s(n) = 2a(n+1) - 2^(n+2) - 2a(n) + 2^(n+1)
a(n+1) = 2a(n) + 2^(n+1),

a(n+1)/2^(n+1) = a(n)/2^n + 1,
{a(n)/2^n}是首项为a(1)/2 = 2,公差为4的等差数列。
a(n)/2^n = 2 + (n-1) = n+1,

a(n) = (n+1)2^n.

追问

为什么最后是a(n+1) = 2a(n) + 2^(n+1)？

追答

a(n+1) = s(n+1)-s(n) = 2a(n+1) - 2^(n+2) - 2a(n) + 2^(n+1),

a(n+1) = 2a(n) + 2^(n+2) - 2^(n+1) = 2a(n) + 2*2^(n+1) - 2^(n+1) = 2a(n) + 2^(n+1)