设已知数列{an}满足a1=1,an－2an-1－2的n-1次方＝0（n∈N*,n≥2）.求证：数列{2的n次方分之an}是等差数列.若数列{an}的前n项和为sn,求sn.别用专业术语

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设已知数列{an}满足a1=1,an－2an-1－2的n-1次方＝0（n∈N*,n≥2）.求证：数列{2的n次方分之an}是等差数列.若数列{an}的前n项和为sn,求sn.别用专业术语
设已知数列{an}满足a1=1,an－2an-1－2的n-1次方＝0（n∈N*,n≥2）.求证：数列{2的n次方分之an}是等差数列.若数列{an}的前n项和为sn,求sn.别用专业术语

设已知数列{an}满足a1=1,an－2an-1－2的n-1次方＝0（n∈N*,n≥2）.求证：数列{2的n次方分之an}是等差数列.若数列{an}的前n项和为sn,求sn.别用专业术语
a(n) = 2a(n-1) + 2^(n-1),
a(n+1) = 2a(n) + 2^n.
a(n+1)/2^(n+1) = a(n)/2^n + 1/2,
{a(n)/2^n}是首项为a(1)/2=1/2,公差为1/2的等差数列.
a(n)/2^n = 1/2 + (n-1)/2 = n/2.
a(n) = n*2^(n-1),
s(n) = a(1) + a(2) + a(3) + ... + a(n-1) + a(n)
= 1 + 2*2 + 3*2^2 + ... + (n-1)2^(n-2) + n2^(n-1).
2s(n) = 1*2 + 2*2^2 + ... + (n-1)2^(n-1) + n2^n.
s(n) = 2s(n) - s(n) = -1 - 2 - 2^2 - ... - 2^(n-1) + n2^n
= n2^n -[1+2+...+2^(n-1)]
= n2^n - [2^n - 1]/(2-1)
= n2^n - 2^n + 1
= 1 + (n-1)2^n

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