Question

已知数列{an}满足a1=1,am+1=2am+1,n∈N*

Answer 1

a(n+1) = 2a(n) + 1,

a(n+1) + 1 = 2[a(n)+1],

{a(n)+1}是首项为a(1)+1=2,公比为2的等比数列.

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

a(n) = 2^n - 1.

b(n) = n[a(n)+1] = n*2^n.

s(n) = b(1) + b(2) + b(3) + ...+ b(n-1) + b(n)

= 1*2 + 2*2^2 + 3*2^3 + ...+ (n-1)*2^(n-1) + n*2^n,

2s(n) = 1*2^2 + 2*2^3 + ...+ (n-1)*2^n + n*2^(n+1),

s(n) = 2s(n) - s(n) = -1*2 - 1*2^2 - 1*2^3 - ...- 1*2^n + n*2^(n+1)

= -2[1+2+2^2 + ...+ 2^(n-1)] + n*2^(n+1)

= n*2^(n+1) - 2[2^n - 1]/(2-1)

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

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