已知数列{an}满足a1=1,am+1=2am+1,n∈N*
问题描述：
已知数列{an}满足a1=1,am+1=2am+1,n∈N*
（1）求数列｛an｝的通项公式（2）设bn=n（an+1）,求数列｛nn｝的前n项和Sn
答
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