Question

设数列an满足a1=1,a(n+1)-an=1/2^n.(1)求数列an的通项公式，（2）令bn=nan，求其前n想和

Answer 1

(1)a(n+1)-an=1/2^n
an-a(n-1)=1/2^(n-1)
......................
a2-a1=1/2^1
累加得 an-a1=1/2^(n-1)+1/2^(n-2)+.+1/2
=1-1/2^(n-1)
an=2-1/2^(n-1)
(2)bn=2n-n/2^(n-1)
令cn=2n,dn=n/2^(n-1)
则cn的前n项和为S1=2+4+.......+2n=n(n+1)
dn的前n项和为
S2=1/2^0+2/2^1+...........n/2^(n-1)
S2/2=1/2^1+2/2^2+..............+n/2^n
两式相减 得 S2/2=1+(1/2+1/2^2+.......+1/2^(n-1))-n/2^n
=1+1-1/2^(n-1)-n/2^n
=2-(1-n/2)/2^(n-1)
因此 S2=4-(1-n/2)/2^(n-2)
故 S=S1-S2=n(n+1)-4+(1-n/2)/2^(n-2)