Question

a1=1,an+1=2an+2^n 设bn=an/2^n-1 1证明bn是等差数列 2求an前n项和sn

Answer 1

n = an/2^(n-1)

b = a/2^(n-2)

bn - b

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

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

把 已知条件 a = 2an+2^n 即 an = 2a + 2^(n-1) 代入上式

bn - b

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

= 1

因此 bn 是等差数列

b1 = a1/2^(1-1) = 1/1 = 1

bn = n

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an/2^(n-1) = n

所以

an = n * 2^(n-1)

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Sn = a1 + a2 + a3 + …… + a + an

= 1 + 2*2 + 3*2^2 + …… + (n-1)*2^(n-2) + n * 2^(n-1)

2Sn = 2 + 2*2^2 + 3*2^3 + …… + (n-1)*2^(n-1) + n * 2^n

两式子相减, 把 2的乘方相同的相合并在一起

2Sn - Sn = Sn

= -1 + (1-2)*2 + (2-3)*2^2 + (3-4)*2^3 + …… [(n-1) -n]*2^(n-1) + n*2^n

= n*2^n - [ 1 + 2 + 2^2 + …… 2^(n-1)]

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

= n * 2^n - 2^n + 1

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