已知数列{an}满足a1=2,a(n+1)=an-1⼀n(n+1）。（1）求数列的通项公式（2)设bn=nan*2^n,求数列的前n项和

(1)
a(n+1)=an -1/[n(n+1）]
a(n+1) - an = -1/[n(n+1)]
= 1/(n+1) - 1/n
an - a1 = 1/n -1
an = 1/n +1
(2)
bn = nan . 2^n
= n(1/n+1). 2^n
= (1+n) 2^n
= 2^n + n.2^n
summation bn
= summmation { (n+1)2^n}

consider
1+x+x^2+..+x^(n+1) = (x^(n+2)-1)/(x-1)
1+2x+..+(n+1) x^n
= [(x^(n+2)-1)/(x-1)]'
= [(n+1)x^(n+2) -(n+2)x^(n+1) +1]/(x-1)^2

2x+3x^2+..+(n+1)x^n
= [(n+1)x^(n+2) -(n+2)x^(n+1) +1]/(x-1)^2 -1

put x= 2
2.2^1+2.2^2+...(n+1).2^n
= (n+1)2^(n+2) - (n+2)2^(n+1) +1 -1
= n.2^(n+1)
summation bn
= n.2^(n+1)